son Regulator Series, No. 




TtlB 

Ingineei's Epitome 



PUBLidHED BY 

HE MASON REGULATOR CO. 

BOSTON, MASS. 



Mason Regulator Company Series, No. 3 
^t^3l THE 

ENGINEER'S Epitome 



A COLLECTION OF 



FIGURES, FACTS, AND FORMUL/E 

For Engineers, by an Engineer of Thirty Years' 
Experience 

^y BY 

" N. J. SMITH, Hartford, Conn. 



PRICE, FIFTY CENTS 



PUBLISHED BY 

THE MASON REGULATOR COMPANY 

Boston, Mass. 



\^'^^v\ 






COPYRIGHTED 

1892 

BY THE MASOK REGULATOR COMPANY 
BOSTON, MASS. 



d^ 



^^- S^3^'0i/ 



Press of Geo. H. Ellis, 141 Franklin Street, Boston. 



PREFACE. 



Nearly two years ago we bought the edition of a book 
on Engineering which, until that time, had had a limited 
circulation. It was an excellent work, and we thought 
ought to be in the hands of every engineer. As such 
works ordinarily are sold at high prices, we placed this 
on the market and advertised it for sale at a price which 
would only cover the cost of the book and the expense 
of handling it. The sale at once exceeded our expecta- 
tion, and a second edition was published. We then 
bought an edition of another book of interest to engi- 
neers, which also met with a large sale. The present 
work is from the pen of N. J. Smith, of Hartford, Conn., 
an engineer of over thirty years' experience. It has cost 
somewhat more to publish, owing to the mathematical 
work contained therein ; but we trust will be found to be 
of sufficient value to the engineer to warrant the increase 
in price. In regard to the pages descriptive of our goods, 
to be found in the latter part of the book, we would say 
that so many of those ordering our previous publications 
asked us to enclose our catalogue that we now add it in 
a form convenient for future reference. 

MASON REGULATOR CO. 

10 Central Street, Boston, Mass. 
Oct. 1, 1892. 



INDEX. 



PAGE 

Fractions 1 

To read Decimals 1 

To write Decimals 2 

Addition of Decimals 3 

Subtraction of Decimals 4 

Division of Decimals 6 

Powers and Roots 9 

Indices and Exponents 9 

Square Root . 11 

Properties of Steam 16 

To find the Units of Heat 17 

Temperature of Steam 18 

Latent Heat of Steam 19 

Total Heat of Steam 20 

Temperature and Vacuum 21 

Temperature in a Partial A^acuum 22 

Temperature from a Vacuum Gauge 22 

Latent Heat in Vacuum 22 

Latent Heat from Vacuum Gauge ....... 23 

Total Heat in Vacuum 23 

Total Heat from Vacuum Gauge 23 

Volume, Weight, and Velocity of Steam ... 25 

Relative Volume of Steam 25 

Weight of Steam 26 

Volume of Steam 27 

Equivalent of Steam in Water 27 

Velocity of Steam 29 

Velocity of Steam in Vacuum 29 

Volume of Steam in Partial Vacuum 30 

Weight of Steam in Partial Vacuum ...... 31 



VI INDEX. 

PAGE 

Boilers 33 

Bursting Pressure 33 

Shearing Strength of Rivets 34 

Diameter of Rivets 36 

Area of Rivets 40 

Percentage of Strongest Joints 41 

Segments of Circles 43 

Thickness of Sheet 44 

Working Pressures 47 

To find Chord from Height 50 

Area of Safety Valve 53 

Weight on Safety Valve 54 

Rise of Pressure 59 

Relative Evaporation 60 

Tests for Moisture in Calorimeter 61 

Chimneys 65 

Temperature in Flue 65' 

Velocity of Draft 66 

Consumption of Coal 68 

Horse-power of Chimney 69 

Area of Chimney 70 

Correct Height of Chimney 70 

Horse-power of Engines 74 

Horse power of Compound Engines 77 

Condensation 77 

Point of Cutoff 78 

Thrust on Propeller 82 

Salt Water Feed and Jet Condensers 83 

Slide Valves 86 

Angular Advance of Eccentric 88 

Inside Lap and Release 89 

Width of Port Opening 92 

The Proportions of Slide Valves 95 

Relative Positions of Piston and Crank 97 

Strength of Principal Working Parts of Engine 99 



INDEX. Vll 
PAGE 

Water: Velocity of Discharge 102 

Weight of Water in Pipes 106 

To measure Kanning Water 109 

Horse-power of Water Wheels . . = o „ . . . Ill 
Horse-power of Pumps ...........112 

Diameter of Suction Pipe 11.5 

Belts : Their Priction, Speed, Horse-power, and 

Arc of Contact 117 

Governors 125 

Height of Balls 126 

Centrifugal Porce 126 

Pipe Surface for Heating , , . . , 128 

Weight of Shafts 128 

Cubical Contents of Wedges, Shafts, Segments, 

AND Zones , . 129 

Area of Segments , 130 

Sine of Angles Less than Pifty , 1.31 

Length of Ctcloidal Circles ........ 131 

Elevation of Outer Rail in Curves 132 

Minimum Tractive Porce of Locomotives . . . 133 

Salt Water and its Properties 134 



THE ENGINEER'S EPITOME. 



FRACTIONS. 



Fractions are of two kinds ; namely, vulgar and 
decimal. 

A vulgar fraction is one whose numerator and denom- 
inator may be any number. 

A decimal fraction is one whose denominator is al- 
ways 1, with a cipher or a number of ciphers annexed. 
In practice, the denominator of a decimal fraction is 
indicated by the position of the numerator relative to 
the decimal point. Thus | is a vulgar fraction, j^-^ is a 
decimal fraction, represented as a decimal in this form: 
.5, the denominator, 10, being understood. 

Since numbers increase in value ten times by moving 
them one place to the left, it is evident that they de- 
crease ten times by moving them one place to the right. 
The value of a decimal fraction may be shown in this 
manner : — 



Five tenths 


• • -fo 


.5 


Five hundredths .... 


■ TOO 


.05 


Five thousandths . . . 


• TOOO 


.005 


Five ten thousandths . . 


Toooo 


.0005 


Five hundred thousandths 


Too'ooo 


.00005 



A mark called a decimal point is placed at the left 
hand of decimals, or between the whole number and the 



Z THE ENGINEER S EPITOME. 

decimal. The first figure to the right of the decimal 
point is tenths, the second hundredths, the third thou- 
sandths, etc. 



H S H H H S W 

12 3 4 5 6 7 

To READ Decimals. — Read the significant figures as 
whole numbers, and give the name of the place corre- 
sponding to the distance of the right-hand figure from 
the decimal point. Thus .0034: is read thirty-four i;en 
thousandths, the last right-hand significant figure, 4, 
being the fourth from the decimal point, or in the ten 
thousandths place. The number .00009 is read nine 
hundred thousandths, the figure 9 being the fifth figure 
from the decimal point, or in the hundredth thousandth 
place. 

To write Decimals. — Write the number as a whole 
number. Then prefix, as is necessary, one or more 
ciphers, to remove the right-hand figure to the required 
denomination. 

Examples. — "Write four ten thousandths. Ten thou- 
sandths being the fourth denomination from the decimal 
point, the figure 4 must also be the fourth from the deci- 
mal point; and three ciphers must be prefixed, thus 
.0004. . Write ninety-seven millionths. Millionths being 
the sixth denomination from the decimal point, four 
ciphers must be prefixed. Write the following numbers 



THE ENGINEER S EPITOME. 6 

Four hundredths, 04 

Ninety-five thousandths, 095 

Seven and four hundred and ninety-three ten 

thousandths, 7.0493 

The last number is called a mixed number, the figure 
7 being the integral number and .0493 the decimal. 

Addition of Decimals. — Write the numbers so that 
the decimal points will be arranged in a vertical row, the 
integral numbers, if any, to the left of the decimal point ; 
and the figures of the same denomination under each 
other ; add as in whole numbers, and place the decimal 
point under the points above. 

Examples. — Add the following numbers : — 

Seven and four thousandths . . . 7.004 

Nine hundredths 09 

Sixty-five and two ten thousandths 65.0002 

Forty-five hundredths 45 

72.5442 

Ansiver. — Seventy-two and five thousand four hundred 
and forty-two ten thousandths. 

Add the following numbers : Twenty-five and seventy- 
three hundredths, eighty-five ten thousandths, ten and 
forty-two hundred thousandths, three and eighty-five thou- 
sandths. Answer. — 38.82392, or thirty-eight and eighty- 
two thousand three hundred and ninety-two hundred 
thousandths. 

Ciphers placed at the right hand of decimals do not 
alter their value, as shown thus : yq%%%% or nine thou- 
sand five hundred thousandths. This is shown by 
dividing both numerator and denominator of the vulgar 
fraction by 100, which does not alter the value of the 
fraction. It will then stand thus : ■^q%%, or ninety-five 
thousandths. It might be desirable at times to fill out 



4: THE EXGINEER S EPITOME. 

the places at the right with cipher?, for symmetry of ex- 
pression or to aid the eye in adding; but the result will 
remain the same. 

To SUBTRACT DECIMALS. — Place the numbers of the 
same denomination under each other, and subtract the 
less number from the greater. Place the decimal point 
under the points above. Rules for subtraction say. Place 
the less number under the greater ; but, in performing 
mathematical operations, it very often saves space and 
rewriting to place the greater number under the less and 
subtract the upper from the lower. 

Exajnples. — From one and one thousandth subtract 
five hundred and forty-nine ten thousandths. 

1.001 
.0549 
.9461 

Answer. — Xine thousand four hundred and sixty-one 
ten thousandths. 

From four hundred and ninety-two and sixty-five hun- 
dredths subtract forty and ninety-four ten thousandths. 

492. G5 
40.0094 



Answer. — Four hundred and fifty-two and six thou- 
sand four hundred and six ten thousandths. 

The weight of a cubic foot of steam at a certain press- 
ure is three hundred and seventy-six ten thousandths of 
a pound. At another pressure it is five hundred and 
ninety-two thousandths of a pound. What is the differ- 
ence in the weight? Answer. — .55-1-1, or five thousand 
five hundred and forty-four ten thousandths of a pound. 

]Multiplication of decimals is performed as multiplica- 



THE ENGINEER S EPITOME. 

tion of whole numbers. Point off as many places for 
decimals in the product as there are in the numbers to 
be multiplied 

Examples. — Multiply ten and six hundredths by sixty- 
three thousandths. In multiplication of decimals it is 
not necessary to place the decimal points under each 
other, as in addition and subtraction ; but the right-hand 
significant figures must be placed under each other and 
multiplied as usual. 

10.06 
.063 
3018 
6036 
Answei' .63378 
There are five decimal places in the numbers to be mul- 
tiplied, so there must be five in the product. Multiply 
forty-seven ten thousandths by sixty-three hundred thou- 
sandths. 

.0047 
.00063 



.000002961 

In this example, as there are nine decimal places in 
the numbers to be multiplied, there must be nine decimal 
places in the product. Therefore, five ciphers must be 
prefixed to the product, making it read two thousand 
nine hundred and sixty-one billionths 

It is very often convenient, when the space at the left 

I of the factors is small, instead of multiplying by the 

right-hand figure first, to multiply by the left-hand 

figure first, and extend the partial products to the right 

Instead of to the left. 

Examples. — Multiply fifty-two and four hundred and 



b THE ENGINEER S EPITOME. 

seventy-five thousandths by twelve and eighty-four 
hundredths. 

In this example, instead of multiplying by the right- 
hand figure first, multiply by the left-hand figure, which 
is 1. Xext multiply by the figure 2, and so on, until all 
the numbers have been multiplied, placing the products 
one place to the right for every figure in the multiplier. 
Thus : — 

.52.47.5 
12 84 
5247.5 
104950 
419800 
209900 



Ansiver 673.77900 

There are in this example five decimal places. There 
should also be five decimal places in the product. Two 
figures of the product are ciphers ; but as they, when at 
the right of the decimal, do not alter its value, the an- 
swer reads six hundred and seventy-three and seven 
hundred and seventy-nine thousandths. 

If steam rushes through a pipe with a velocity of one 
thousand five hundred and ninety-seven and thirty-seven 
hundredths feet per second, how far will it go in one 
hundred and sixty-seven hundredths seconds? Ansiver. 
— 160807.2379 feet. 

Division of Decimals. — Dividing by the significant 
figures is performed as division of whole numbers. The 
number of decimal places in the divisor and quotient 
must equal the number of decimal places in the divi- 
dend, or the difference between the number of decimal 
places in the dividend and divisor equals the number of 
decimal places in the quotient. If the divisor is extended 
beyond the original figures, every cipher thus annexed 



THE ENGINEER S EPITOME. 7 

must be considered as decimal figures of the dividend. 
The quotient can be extended to any degree of accuracy. 
If the dividend is a whole number and the divisor a 
decimal, annex as many ciphers to the dividend as there 
are figures in the divisor. The quotient will be a whole 
number till the divisor is extended to the last of the an- 
nexed ciphers. If more ciphers are annexed, the con- 
tinued figures of the quotient will be decimals. If the 
divisor is a whole number and the dividend a decimal, 
the quotient will contain as many decimal figures as 
there are figures in the dividend. 

Examples. — Divide forty-five hundredths by six and 
eighteen ten thousandths. In this example the divisor 
contains five figures, a unit and four decimals, and the 
dividend but two figures, which are decimals. Now, as 
annexing ciphers to the right of the decimal does not 
alter its value, annex four ciphers to the right of the 
dividend, making six figures, which is the least number 
that will contain the divisor. Place the figure 7, which 
denotes the number of times the divisor is contained in 
the dividend, in the quotient. As there are but four 
decimal places in the divisor and six in the dividend, 
there must be two in the quotient. Therefore, a cipher 
should -be prefixed to the quotient. The position of the 
decimal point being thus ascertained, the division can be 
carried to any degree of accuracy. The answer, approx- 
imately, is seven hundred and forty-nine ten thou- 
sandths. 

6.0018).450000(.0749 Answer. 
420126 

298740 

240072 
586680 
540162 



O THE ENGIXEEII S EPITOME. 

Divide four Imnclred and seventj-six and five tentlis 
by eighty-four thousandths. The dividend has but one 
decimal place, and the divisor three. Annex two ciphers 
to the dividend, making the number of decimal places in 
the dividend and divisor equal. Divide all six figures 
in the dividend before placing the decimal point. Then 
annex other ciphers, and carry the division as far as 
necessary. 

.084)476.500(5072 Answer. 
420 
565 
504 
610 
588 
220 «- 
168 ■ 
52 

The pressure of steam on an area of one hundred and 
twenty-five and forty-three hundredths square inches is 
eight thousand and thirty-seven thousandths pounds. 
What is the pressure on one square inch? Answer. — 
63.78 pounds approximately. 

These questions have been written in full, to give the 
learner practice in reading and writing decimals by com- 
paring the numbers as expressed by written words and 
figures. 



THE ENGINEER S EPITOME. 



POWERS AND ROOTS. 



A power of a quantity is the continued multiplication 
of the quantity by itself. The number of times the 
quantity is taken as a factor is shown by a figure placed 
at the upper right-hand corner of the quantity, called an 
index or exponent of the power; Thus 3'^. Two is the 
index of the quantity three, and signifies that the quan- 
tity three is twice a factor, or multiplied by itself once, 
as 3- = 3x3=:9. 4^ signifies that the quantity four is 
three times a factor, or multiplied by itself twice, as 
43=4x4x4r=64. 

The index or exponent of a power can be either as a 
whole number, as 8"^, a fraction, as 8*, a decimal frac- 
tion, as 8.45, or a minus index, as 8— |. When the index 
of the power is fractional and the numerator is one, it is 
usually called a root. A root differs from a power in- 
asmuch as it is one of the equal factors of the same num- 
l3er as the denominator of the index of the power, which, 
multiplied together, will produce the quantity. Thu-s 
64- = 8, eight being one of the two equal factors, the 
product of which is sixty-four ; also 125^ = 5, five being 
one of the three equal factors, the product of which is 
one hundred and twenty-five. When the numerator of 
the index of the power is any other figure than one, as 
8i, it shows that the quantity eight must be multiplied 
by itself, and the fifth root of the prod act taken, or the 
:fifth root of the quantity eight taken, and this root mul- 
tiplied by itself. 

Generally, the first process will be the best. The 
minus index denotes the reciprocal or one divided by 



10 THE exgineek's epitome. 

the result of the operation, or the quantity denoted by 
the index. Thus 5— -rr^-rzJ^. The root of a quan- 
tity is most generally shown by a mark called a radical 
sign. Thus aI with a figure denoting the root re- 
quired placed above it. Thus a/27 signifying that the 
cube root of twenty-seven is required. When no figure 
is placed above the radical sign, the figure 2, or the 
square root, is always understood. 

The relative value of the index is shown thus : — 
G4i 643 64-2 641 q^i 54! 64* 54 J 54^ etc. 



The last three numbers are most often expressed thus, 
-J 64, -v/ 64, Ai 64. The power of any number is 1. 
The fourth power is most easily found by squaring the 
square, and the sixth power by squaring the cube, etc. 
The fourth root is easiest found by extracting the square 
root of the square root, the sixth root by extracting the 
cube root of the square ruot. 

By multiplying the power denoted by any index by 
the power denoted by any other index, the power de- 
noted by the sum of the indexes is obtained. By ex- 
tracting the roots denoted by the factors of the index of 
the required root of any quantity, the required root of 
the quantity is obtained. Thus a / a = -W of a/ a =: 
A/' of A/ a = -W of a/ of a/ of a/ a ; and a/ a = 
■\i of '^ a = a/ of a/ a = a/ of a/ of a,' a. Ex- 
pressed in the fractional form thus: a^ of a*; a^ of a^; 

or thus : i / a/ a ; or thus : a/ ^K ©tc. 



THE engineer's EPITOME. 11 

When the product of the indices of the partial roots 
equal the index of the required root, ten times the square 
root of any number is the square root of one hundred 
times that number. Thus the square root of .5 = .707 ; 
the square root of 50, or 100 times .5, is 7.07. The 
square root of 64 is 8. The square root of 6100, which 
is 100 times 64, is 80. 

As the square and square root form the base from 
which more mechanical rules can be formed than any 
other, I give the following rule for extracting the square 
root. 

Point the number into periods of two figures each by 
placing a small mark over the first unit or right-hand 
figure, the third, the fifth, and so on. A period, counting 
from the left hand, ends with the figure under the mark. 
If the number of the figures are even, the left-hand 
period will consist of two figures ; if odd, of but one. If 
the number contains a decimal, continue the marks over 
every other figure toward the right. If the number is 
a decimal, assume a unit figure, and point off from that 
toward the right. Find the largest number which, when 
multiplied by itself, gives a product which is equal to 
or less than the left-hand period ; place it to the right 
of the number whose root is to. be found, in the manner 
of a quotient; subtract its product from that period, 
and to the right of the remainder, if any, bring down 
the next period, which call a new dividend. To the left 
of the new dividend place twice the amount of the 
quotient figure already found : annex, or suppose to be 
annexed, a cipher to this last amount. Using this as a 
trial divisor, find the number of times the trial divisor 
is contained in the new dividend ; place the figure denot- 
ing the number of times, in the quotient, also in the 
trial divisor in the place of the supposed cipher. Multi- 



12 THE EXGIXEER's EPITOME. 

ply the complete trial divisor by the last figure in the 
quotient, and subtract the product from the new divi- 
dend. To the right of the remainder, if any, bring down 
the next period for another dividend; double the last 
figure in the complete divisor, bring it, with the rest of 
the divisor, down to the left of the new dividend ; annex, 
or suppose to be annexed, a cipher, and proceed as be- 
fore. Continue this operation ilntil all the periods are 
brought down. A few examples will show the process 
more plainly than a rule. 

Example^. — Find the square root of 23796. Pointing 
off the periods, we find that the first or left-hand period 
contains but one figure. The largest figure which, mul- 
tiplied by itself, is equal to or less than 2 is 1. Multiply 
1 by itself, and subtract the product from the figure of 
the first period. The remainder is 1. Bring down the 
next period, double the quotient figure already found, 
and place it to the left of 137, in the manner show^n in 
the illustration. After supposing a cipher to be annexed, 
the trial divisor is 20, which is contained in 137 six 
times ; but, as the figure 6 will take the place of the 
supposed cipher, making the trial divisor 26, which is 
contained in 137 only five times, 25, therefore, must be 
the trial divisor. Place 5 in the quotient, and by it 
multiply 25, making 125, which subtract from 137 ; to 
right of the remainder, bring down the next period. 
Add the last figure in the quotient to the former divisor, 
for a new divisor. After the supposed cipher is annexed 
to 30, find the number of times the trial divisor is con- 
tained in the new dividend and place that figure in 
the quotient and in the divisor in the place of the 
assumed cipher. ISIultiply the divisor by the last figure 
in the quotient, and subtract the product from 1296. 
Now, all the periods have been brought down and di- 



THE engineer's EPITOMK. 13 

Tided, and there is still a remainder, w hicli shows that 
23796 is not an exact square ; that is, the root 154, mul- 
tiplied by itself, will lack SO of making 23796. By an- 
nexing periods of ciphers, a decimal approximation can 
be obtained. Generally, three or four decimal places 
make an approximation close enough for all practical 
purposes. There must always be as many periods in the 
root as there are periods in the number. 

23796(154 Answer. 
1 
25)137 
125 
304) 1296 
1216 



Find the square root of 1002001. Here, as before, the 
number of figures being odd, there is but one figure in 
the first period. The largest number which, multiplied 
by -itself, is equal to or less than 1 is 1. Place 1 as the 
first quotient figure. Multiply 1 by itself, and subtract 
the product from the first period : nothing remains. 
Bring down the next period, which is two ciphers, mak- 
ing three ciphers. Double the last figure of the quotient, 
and place it at the left of the new dividend. Suppose a 
cipher to be annexed, making 20 for the divisor, which 
is not contained in the dividend. Place a cipher in the 
quotient, also in the place of the assumed cipher. Bring- 
ing down the next period, we find that the divisor, with 
its assumed cipher, is not contained in the new dividend. 
As before, a cipher must be placed in the quotient and 
in the place of the supposed cipher. Bring down the 
next period, placing it to the right of the former one; 
the trial divisor, with the assumed cipher, is contained 
in the new dividend once. Place the figure 1 in the 
quotient and in the place of the assumed cipher. Multi- 



11 TIIK engineer's epitome. 

ply the divisor by the last figure in the quotient, and 
subtract the product from the dividend. The result 
shows that 1001 is the exact square root of 100'2001. 

1002001(1001 Answer. 
_l 
200]y0002001 
2001 

Find the square root of the decimal .0527. The firso 
period will include the first two figures at the right of 
the decimal point, and the second the last two figures. 
In decimals, the first period must always contain two 
figures. 

The largest figure which, multiplied by itself, is equal 
to or less than the number contained in the first period 
is 2. The operation can now be carried on as in previous 
examples. In this case, two ciphers have been brought 
down and the quotient extended. As it is not intended 
to make any closer approximation, the last multiplica- 
tion is not performed. 

.0527(.2296 Answer. 
J- 
42)127 
_84 
449) 4300 
4041 
4586)"25900 

Find the square root of .005847. 

.005847 (.07 646 Answer. 
49 
146)~947 
876 
1.524) 7100 
6096 
1528)100400 



THE engineer's EPITOME. 15 

If at any time the dividend will not contain the trial 
divisor (see second example), another period must be 
brought down : if this will not contain it, another, and 
so on ; but for every period thus brought down that does 
not contain the divisor a cipher must be annexed to the 
divisor. 

Although a lengthy explanation, it will be seen that 
the operation is very shoit. A little practice will enable 
one to extract the square root of a, number, with very 
few figures. 



16 THE engineer's EPITOME. 



PROPERTIES OF STEAM. 



The most useful properties of steam are its elasticity, 
capacity for conveying heat, and its facility of condensa- 
tion. 

It requires 180.5 units of heat to raise the temperature 
of one pound of water 180 degrees, — i.e., from the freez- 
ing to the boiling point, or 188.5 pounds of water at 
maximum density one degree, — because the specific heat 
of water increases with the temperature. When the 180 
degrees of heat have been added, it will become no hotter, 
but commence to boil, and continue at that heat until 
evaporated. If the evaporation should take place in a 
closed vessel, the vapor would be compressed, a pressure 
would be exerted in the vessel, and the temperature of 
the water and vapor would rise. When the temperature 
of the steam, in relation to the pressure, is such that, if 
any heat is withdrawn, a part of the steam would return 
to water, the steam is said to be saturated. If to such 
steam additional heat is added, the steam is said to be 
superheated. 

Steam removed from direct connection with water can 
be heated to almost any extent. Water in the process 
of vaporizing or changing to steam gives to the steam an 
amount of heat not recognized by a heat gauge, which 
is called latent heat, which is given back or out when 
the steam is condensed. Thus one pound of water at 
212° changed to steam at 212°, and this steam being all 
condensed in 5.6 pounds of water at 32°, will raise the 
temperature of the whole to 212°. 



THE engineer's EPITOME. 17 

The sum of the heat due to specific increase of tem- 
perature of the water, the sensible heat of the steam due 
to pressure, and the latent heat due to pressure, equal 
the total heat of the steam. 

To FIND THE Units of Heat required to raise the 
temperature of one pound of water from 32° to tempera- 
ture due to pressure, use the following rule : From 165 
subtract the gauge pressure until the gauge pressure 
equals 10. To the remainder add 17^ times the square 
root of the gauge pressure, and from this sum sub- 
tract .193 of the gauge pressure. 

Formula (for gauge pressure above 10) : 



155+17^^' g- 



- .193 



A short rule : To 164 add 14^ times the square root of 
the gauge pressure. 

Examples. — How many units of heat will be required 
to raise one pound of water from 32° to temperature 
corresponding to 4 pounds' gauge pressure ? 

Example No. 1. 

165 — 4 = 161 

17iX 2-34.5 

161+34.5 = 195.5 

.193 of 4 = .772 

195.5 — .772 = 194.728 units of heat. Answer. 

Example Xo. 2. — Find the number of units of heat 
required t - raise water from 32° to 75 pounds' gauge 
pressure. 

•a/75~(8.66 8.66 



17 



166)1100 
996 
1726)10400 



18 THE kngineek's epitome. 

Example No. 3. — Find the number of units required 
to raise water from 32° to 150 pounds' gauge pressure. 
(By short rule.) 

)150(12.24 X 14i--- 17.5.44 Atisiver. 
1 
22)0.50 
44 
242)'~l50O 
484 
2444)11600 
9776 

To FIND THE Units of Heat required to raise the tem- 
perature corresponding to one gauge pressure to that of 
another, use the following rule : From 17^ times the 
difference between the square roots of the pressures sub- 
tract .193 times the pressures. 

Formula: 17:|-(-\/'g — -J g ) — .193 (g — g), in which 
g and g are the pressures. 

Short rule: 14 1- (-\Jg — -yj g V which signifies multi- 
plying the difference between the square roots of the 
pressures by 14^. These '' short rules " are for gauge 
pressures below 175 pounds. 

Examples. — Find the number of units of heat required 
to raise the pressure from 64 to 121 gauge. 

(By short rule.) 

^/64^8 

-i/T2T= 11 

11—8 = 3 

3 X 141 = 43 units of heat. Ajiswer. 

To find the Temperature of Steam, use the follow- 
ing rule. From 197 subtract the gauge pressure until 
the gauge pressure equals 10. To the remainder add 17 



THE engineer's EPITOME. 19 

times the square root of the pressure. From this sum 
subtract ^ the pressure. 

Formula (for pressures over 10) : (187 + 17 -a/ g ) — i g. 
Short rule : To 199 add 14 times the square root of 
the pressure. 

What is the temperature of the steam when the gauge 
shows 100 pounds' pressure ? 
^'Too= 10 
10 X 14= 140 
140 + 199 = 339° A?7swer. 

What is the temperature of the water in the boiler 
when the gauge shows 60 pounds' pressure ? 

^60^ 7.74 

7.74 X 17-- 131.58 

187 + 131.58 = 318.58 

iofGO= 12 

318.58 — 12 = 306.58° Ansiver. 

A logarithmic rule is given for those who wish. It is 

simplified from D. Y. Clark's formula, but will not vary 

more than i degree from the original. 

2.940 

Formula: ^ ^ "\ ^o72r=T. "p" signifies ab- 

6.2 + Log. p ^ ^ 

solute pressure. When the absolute pressure is less than 

1, the Log. becomes minus, and must be added to 6.2. 

To FIND THE Latent Heat of Steam. — Rule: To 
979 add |- the gauge pressure, until the pressure equals 
10. From this sum subtract 12 times the square root 
of the pressure. To the remainder add ^ of the pressure. 

Formula : (984 — (12 x ^J~^)) + i g- 

Short rule : From 977 subtract 10 times the square 
root of the pressure. 

Example. — Find the latent heat at 80 pounds' gauge 
pressure. 



.20 THE engineer's EPITOME. 

V'80l= 8.94 

8.94 X 12 = 107.28 

984 — 107.28 = 876.72 

iof80=10 

876.72 + 10 = 886.72° Ansioer. 

Find the latent heat at 6 pounds' pressure. 
979 + 3 = 982 
-yJQ^ 2.44 
2.44X12 = 29.28 
982 + 29.28 = 952.72 
■J- of 6 = f, or .75 
952.72 + .75 = 953.47 Answer. 

To FIND THE Total Heat.— Rule: To 1172 add 5 
times the square root of the gauge pressure. From this 
sum subtract yL the pressure. 

Formula : 1172 + (5 ->^/"g~) — ^\ g. 

Short rule : To 1173.5 add 4^ times the square root of 
pressure. 

Examples. — Find the total heat, the gauge pressure 
being 95 pounds ; also the latent heat, Ihe sensible heat, 
and the units of heat required to raise the water from 32°. 

(By short rules.) 

■yJ95l= 9.74 

9.74X4^ = 42.2 

1173.5 + 42.2 = 1215.7 = total heat. 

-yJ95~= 9.74 

9.74 X 10 = 97.4 

977 — 974 = 8796 = latent heat. 

■yJ9bl= 9.74 

9.74 X 14 = 136.36 

199 + 136.36 = 335.36 = sensible heat. 

-^95^9.74 

9.74X14^=139.6 

1 64 + 139.6 = 303.6 = units of heat. 



THE engineer's EPITOME. 21 



TEMPERATURE IN VACUUM. 



The air pressure in a perfect vacuum is 14.7 pounds, 
when the mercury in the barometer tube stands 30 
inches high, or one pound for every 2.04 inches. Any 
variation in the height of the column of mercury in- 
dicates a corresponding variation in the atmospheric press- 
ure. The pressure exerted inside a closed vessel from 
a perfect vacuum is called the absolute pressure, and is 
the atmospheric pressure due to the height of the column 
of mercury plus the pressure shown by a pressure gauge. 

The units of heat required to raise one pound of water 
from 32° to temperature denoted by absolute pressure, 
less than 14.7 pounds, is practically the temperature of 
the water, less 32°, as the gain is but one-half a unit 
more than the gain in temperature from 32° to 212°. 

Water placed in an air-tight vessel until the vessel is 
partly full, with the air removed to an absolute pressure 
of one pound, will boil or vaporize at a temperature of 
102°. If the heat is continued, the vapor will be com- 
pressed; and the temperature will rise until 212° is 
reached. Now, if the vessel is opened, there will be a 
balance of pressure of vapor inside and of air outside, or 
the vacuum will be destroyed, and any additional heat 
will cause the vapor to be expelled into the air; but, if 
the vessel remains closed, we have the conditions pre- 
viously named. If the vessel is entirely filled with 
water, no steam will be formed ; and the vessel and 
water can be heated to any degree, the only pressure on 



22 THE engineer's epitome. 

the vessel being that due to the expansion of the water, 
the water thus becoming superheated. Water in this 
state, if released from confinement, will immediately 
change to steam. 

To FIND Temperature in a Partial Vacuum. — 
Rule: From the absolute pressure subtract i.9. iMulti- 
ply the square root of the remainder by 27. To the 
product add 117. 



Formula: 117 + 27 ^/p— 1.9. 

To FIND THE Temperature directly from the 
Vacuum Gauge. — Rule: From 26.12 subtract the read- 
ing of the gauge. Multiply the square root of the re- 
mainder by 19. To the product add 116. 

Formula: 116 + 19^/26.12 — r. 

Examples. — The absolute pressure is 9 pounds. What 
is the temperature ? 

9 — 1.9 = 7.1 

'\/7T=^2.66 

2.66 X27 = 71.82 

71.82+ 11" = 188.82 = temperature. 

A vacuum gauge shows 25 inches mercury pressure. 
What is the temperature in the condenser? 

26.12 — 2.5 = 1.12 

-vLT2 = 1.0.5 

1.05 X 19 = 19.95 

19.95 -+ 116 = 135.95 = temperature in condeusor. 

To find the Latent Heat in Vacuum. — Rule: 
From the absolute pressure subtract 1.9. Multiply the 
square root of the remainder by 19. Subtract the prod- 
uct from 1035. 



Formula: 1035 — 19 ^ p — 1.9. 



THE engineer's epitoisie. 23 

To FIND THE Latent Heat directly from the 
Vacuum Gauge. — Rule: From 26.12 subtract the read- 
ing of the gauge. Multiply the square root of the re- 
mainder by 13J. Subtract the product from 1035. 

Formula : 103.5 — 131 ^/ 26.12 — r. 

Examples. — What is the latent heat when the absolute 
pressure is two pounds? 

2 — 1.9 = .l 

.316X19 = 6.004 

1035 — 6.004 = 1028.996 = latent heat. 

The vacuum gauge shows 20 inches. What is the 
latent heat of the vapor in the condenser ? 

26.12 — 20 = 6.12 

-V 6^12 = 2.47 

2.47 X 13i = 1002.07 = latent heat of vapor. 

To FIND THE Total Heat in Vacuum. — Rule : From 
the absolute pressure subtract 1.9. Multiply the square 
root of the remainder by 8. To the product add 1150. 



Formula : 1 1 50 -f- 8 '^' p - 



-1.9 



To FIND THE Total Heat directly from the 
Vacuum Gauge. — Rule: From 26.12 subtract the read- 
ing of the gauge. Multiply the square root of the re- 
mainder by 5|^. To the product add 1150. 
Formula : 1150 + 5^ -^26.12 — r 

Excunples. — The absolute pressure is 10. What is the 
total heat ? 

10 — 1.9 = 8.1 
8T= 8.84 



i" 



2.84 X 8 = 22.72 

22.72+ 1150= 1172.72 total heat. 



24 THE engineer's epitome. 

What is the total heat in condenser, when the gauge 
reads 15 inches? 

26 .12 — 15 = 11.12 
-yl 1.1 2 = .3.33 

3.33 X 5^=18.315 

18.315 + 1150=1168.31.5 total heat. 



THE engineer's EPITOME. 25 

VOLUME, WEIGHT, AND VELOCITY 
OF STEAM. 



The volume of steam is the space it occupies in rela- 
tion to the space occupied by the water from which it 
is formed. One cubic foot of water, changed to steam 
at atmospheric pressure? will occupy 1646 cubic feet of 
space, or a ratio of 1646 to 1. If the pressure falls below 
the atmospheric, this ratio increases very rapidly. If 
more steam is formed and the pressure raised, the ratio 
continually decreases. 

The weight of the steam is equal to the weight of the 
water from which it was formed, and is equal, space for 
space, to the weight of the water divided by the relative 
volume ; i.e., the weight of one cubic foot of steam at 
any pressure is equal to the weight of one cubic foot 
of water divided by the relative "volume of the steam at 
that pressure. The flow or velocity of steam follows the 
same, and is subject to the same laws as those of the flow 
of other fluids. 

To FIND THE Relative Volume of Steam, also 
Gauge Pressure. — Rule: Divide 81, plus the square 
root of the gauge pressure, by .05, plus -g^Q- of the gauge 
pressure. 

81 + A/'~g^ 

Formula : -— -^-^ — (-gi^ expressed as a decimal 

.05 -|- .OOoo g 

is .0033). 

Examples. — The pressure on the steam gauge is 90 

pounds. How many times is the bulk of water from 

which the steam was formed expanded? 



26 THE engineer's epitome. 

-y 90= 9.47 

81+9.47 = 90.47 

.05 + ^ip of gauge pressure = .35 

90.47 4- .35 = 258.48 Answer. 

What is the relative volume of steam when the gauge 
pressure is 30 pounds? 

^30 = 5.47 

81 4- 5.47 = 86.47 

3^0 of 30= 1 

.05 + .1 =.15 

86.47 -r. 15 = 576.46 = relative volume of steam Answer. 



Another rule is as follows : Multiply the fourth root of 
the absolute pressure by 2820. To the product add 
18800, Divide the sum by the absolute pressure plus 
.05. To the quotient add the fourth root, and subtract 
10. This rule gives very correct results, from 1 pound 
absolute upward. 

Example (illustrating the latter rule). — What is the 
relative volume of the steam, the gauge pressure being 
65 pounds? The absolute pressure in this example is 
79.7 pounds, and the fourth root equals the square root 
of the square root. 



V: 



79.7 = 8.92 
^ = 2.98 = fourth root. 



2,98 X 2820 = 8403.6 

8403.6 + 18800 = 27203.6 

79.7 + .05 = 79.75 

27203 6-1-79.75 = 341.11 

341.11 +2.98 = 344 09 

344.09 — 10 = 344.09 = relative volume of steam Answer. 



To FIND THE Weight of a Cubic Foot of Steam. 
Rule: To 3.125 add .208 times the gauge pressure, and 



THE engineer's EPITOME. 27 

divide the sum by 81 plus the square root of the gauge 

pressure. 

-r, , 3.12o+.208g 
i orm ul a : — ° 

si+V g 

Example. — What is the weight of a cubic foot of 
steam, the gauge pressure being 50 pounds? 

50 X .208 = 10.4 

3.12.5 + 10.4= 1.3.525 

■^50=- 7.07 

81+7.07 = 88.07 

13.525 -;- 88.07 = .1535 pounds Answe?-. 

To FIND THE Volume of a Given Weight of Steam 
IN Cubic Feet. — Rule : Multiply 81 plus the square 
root of the gauge pressure by the given weight. Divide 
the product by 3.12.5 plus .208 times the gauge pressure. 

Formula : — ^^ J 1 in which w indicates the 

3.125+.208g 
given weight of steam. 

Example. — The steam space in a boiler is 90 cubic 
feet. The gauge pressure is 110 pounds. What is the 
weight of the steam ? 

110 X -208 = 22.88 

3.125 + 22.88 = 26.005 

VlTo = 10.48 

81 +10.48 = 91.48 

26.005 ^91.48= .2842 

.2842 X 90 = 25.578 = weight of steam. 

To find the Cubic Inches of Water to make 

One Cubic Foot of Steam. — Rule: To 86.4 add 5f 

times the gauge pressure, and divide the sum by 81 plus 

the square root of the gauge pressure. 

, 86.4 + 5.75 g 
Formula : — -= — - 

I I 



81+\ g 



2o THE KNGIXEEK S EPITOME. 

Examples. — How many cubic inches of water will make- 
2 cubic feet of steam at a pressure of 70 pounds*^ 
gauge ? 

70 X 5| = 402.5 

86.4 + 402.5 = 488.9 

^^70^ 8.36 

81 4- 8.36 = 89.36 

488.9 ^ 89.36 = 5.47 

5.47 X 2 = 10.94 cubic inches. 

The steam space in a boiler is 64 cubic feet. How 
many cubic inches of water will be required to fill it 
with steam at 121 pounds' gauge pressure? 

121 X5f-= 695.75 
86.4 + 695.75 = 782.15 
-1/121=11 
81 + 11=92 
782.15-^92 = 8.501 
8.501 X 64 = 544.06 cubic inches Answer. 



THE ENGINEER S EPITOME. 



VELOCITY OF STEAM. 



The theoretical velocity of steam is the same as that 
of a body falling freely, by gravity, from a height which 
is such that the weight equals the pressure on the same 
area. Thus, if the pressure of steam is 100 pounds per 
square inch, its velocity into perfect vacuum would be 
that of a body falling freely, by gravity, from a height of 
a column of steam one inch square at relative volume 
corresponding to 100 pounds' pressure that would weigh 
100 pounds. 

The theoretic velocity of steam flowing into the air 
or into steam of less density, is the velocity due to dif- 
ference of the heights of two columns of steam, of the 
density due to the highest pressure and of weights equal 
to each pressure. The velocity of steam in a vacuum 
equals 96 times the square root of the number of cubic 
feet required to weigh a pound, multiplied by the abso- 
lute pressure. 

Tq find THE Velocity of Steam in a Vacuum, 
Gauge Pressure. — Rule: From 13 subtract .005 of the 
gauge pressure. Multiply the remainder by the square 
root of the gauge pressure minus .7. To the product 
add 1890. 



Formula: 1890+(l3 — ,005 g) (^g 



. . . --v 

Example. — What is the theoretic velocity of steam in 
a vacuum when the gauge pressure is 100 pounds ? 



30 THE engineer's EPITOME. 

.005 of 100 = .5 

13 — .5 = 12.5 

-^,100=10 

10 — .7 = 9.3 

12.5 X 9.3 = 116.25 

116.25 + 1890 = 2006.25 A)isiver. 

What woald be the velocity of this steam into the air? 

The formula (13 — -005 gW-v/ g — •") shows the in- 
crease in velocity due to the gauge pressure of steam 
escaping into vacuum, 1890 being the velocity at which 
steam at atmospheric pressure will escape into a perfect 
vacuum. Steam at any absolute pressure less than 14.7 
and greater than 1 will flow into a perfect vacuum with 
a velocity equal to 1747 plus 40 times the square root 
of the absolute pressure. 1747 is the velocity at which 
steam at 1 pound absolute will flow into a perfect 
vacuum. Steam is retarded in its flow through pipes 
by friction, elbows, and valves ; and the pipes should be 
of such size that the steam at the engine should at all 
times approximate very closely to that in the boiler. 

To FIND THE Volume of Steam in Partial 
Vacuum.— Rule: To 20600 add 1000 times the square 
root of 13.7 minus the vacuum, and divide the sum by 
14 7 minus the vacuum. 

20600+1000 a/13.7— V 

I ormu!a : 1 

14.7 — V 
Examples. — What is the relative volume of the steam 
when the vacuum is 13.7 pounds ? 

13.7 — 13.7 = 

OX 1000 = 

20600 + = 20600 

14.7 — 13.7 = 1 

20600 -r 1 = 20600 = relative volume of steam Answer. ' 



THE engineer's EPITOME. 31 

The vacuum is 3 pounds. What is the relative 
vokime ? 

13.7 — 3 = 10.7 
^'1077 = 3.27 

3.27 X 1000 = 3270 00 

20600+3270 = 23870 

14.7—3 = 11.7 

23870 -T- 11.7 = 2040.17 = relative volume of steam Answer. 

To FIND THE Weight of Steam in Partial Vac- 
uum. — Rule: From 14.7 subtract the vacuum, and 
divide the remainder by 332 plus 15 times the square 
root of 13.7 minus the vacuum. 

14.7 — V 
Formula : 



332+15^13.7 — V 
Examples. — The vacuum in the condenser is 11 pounds. 
What is the weight of one cubic foot of the steam? 

14.7—11 =3.7 

13.7 — 11 =2 7 

-v'2?r=1.64 

1.64X15 = 24.6 

332 + 24 6 = 356.6 

3.7 -I- 356.6 = .01037 pounds Answer. 

The capacity of a condenser is 50 cubic feet. It con- 
tains 5 cubic feet of water, which weighs 305 pounds. 
What is the weight of water in condenser when steam is 
entirely condensed ? 

14.7—9.7 = 5 

13.7 — 9.7=4 

^4^2 

2X15 = 30 

332 + 30 = 362 

5 -i- 362 = .01381 

50 — 5 = 45 

.01381 X 45 = .62145 

.62145 + 305 = 305.62145 pounds Ansiver. 



32 THE engineer's epitome. 

To FIND THE Cubic Inches of Water to make 
One Cubic Foot of Steam in Partial Vacuum. — 
Rule : From 14.7 subtract the vacuum. Divide the 
remainder by 12 plus .55 times the square root of 13.7 
minus the vacuum. 

Formula : '- — -, i^zr-. 

12-i-.55-vl3.7— V 

Example. — How many cubic inches of water are re- 
quired to make 1000 cubic feet of steam when the 
vacuum gauge shows 24 inches? 

In these rules, g indicates the gauge pressure and v 
the vacuum, except in those rules in which the results 
are obtained directly from the vacuum gauge, in which 
cases V indicates the reading of the gauge. 

24 -^2.04 = 11.76 
14.7_ 11.76 = 2.94 
13.7 — 11.76=1.94 
A!'l94 = 1.39 
.55 X 1.39 = .764.5 
12. + .7645= 12.7645 

2.94 -^ 12.7645 = 2303 = for 1 cubic foot of steam. 
1000 X .2303 = 230.3 = for 1000 cubic feet of steam. 



THE ENGINEER S EPITOME. 



33 



BOILERS. 




The strain tending to rupture a boiler longitudinally 
is the pressure per square inch, multiplied by the length 
of the boiler in inches, multiplied by the diameter in 
inches. From this it is evident 
that the strain to rupture a strip 
of the circumference of the boiler 
one inch in length is equal to the 
pressure multiplied by the diame- 
ter. This strain is always perpen- 
dicular to a diameter of the boiler. 
Thus, in the cut, where A B is s. 
•diameter, the lines of force tending to rupture the boiler 
are c d, e f, etc. This diameter may be horizontal, per- 
pendicular, or inclined at any degree ; but the lines of 
force that rupture the boiler are always perpendicular 
to it. The force on the part of the circumference B d 
tending to rupture the boiler is the pressure multiplied 
by the part of the diameter B c, and the force on the 
circumference d f tending to rupture the boiler is the 
pressure multiplied by the part of diameter c e, etc. 

Supposing the strip to be one inch long (I use the 
word " long " instead of ''"wide," as the strip is measured 
longitudinally with the boiler), there is at each end of 
the diameter a resisting force due to the strength of 
material of which the boiler is composed ; and each end 



34 THE enginf:er's epitome. 

will have to resist but one-half the rupturing force. 
Hence the rule : Divide the tensile strength of the mate- 
rial by one-half the diameter in inches. But the tensile 
strength of a material is given in pounds per square inch 
of section, hence this rule must be modified as follows: 
Multiply the tensile strength by the thickness, and divide 
the product by one-half the diameter. This rule will give 
the bursting pressure of an inch strip, provided the 
part where the two ends are joined be as strong as the 
rest of the strip ; but this is seldom the case, and prob- 
ably, in boilers, never the case. Again, it is evident 
that, if one inch in length will stand the strain due to 
itself, the whole length of the boiler will stand an equal 
strain on each inch of its length. 

The joints or seams in the sheets of which a boiler is 
made are usually formed by rivets, one, two, or more 
rows being used: the boiler being called single, double, 
or triple riveted, according to the number of rows of 
rivets. 

When the shearing strength of section of rivet is 
equal to tensile strength of section of sheet between 
rivet holes, the strongest joint is obtained. This, in 
a single riveted boiler, is assumed to be .56 of solid 
sheet ; in double riveted boilers, .70 of solid sheet. A 
boiler is allowed by government inspection to carry 
but ^ of the pressure that would cause rupture. A factor 
of safety of 5 is generally considered sufficient for a 
stationary boiler. The strain on the rivets is a shearing 
strain, not a tensile strain, or a strain to cut the rivets 
in two instead of pulling them apart. The English 
Board of Trade rule makes the area of rivet section to 
area of sheet section between rivets as 28 to .22, or 
nearly 22 per cent, greater than sheet section. The 
strength of a riveted seam fur boiler use decreases as the 



THE engineer's EPITOME. 35 

thickness of the sheet increases, being .56 in a single 
riveted seam when the thickness is about -f-^, .70 in a 
double riveted seam when the thickness is about ^, and 
.70 in a triple riveted seam when the thickness is about 
1 inch. 

Now, as the area of the rivet increases much faster 
than the diameter, it is possible to make the riveted 
joint approach much nearer the total strength of the 
sheet by using larger rivets and a greater pitch. Such 
a joint might not do for a boiler, on account of a ten- 
dency to leak between the rivets ; but, for bridge and 
other " dry " work, I would recommend a single riveted 
joint of much greater percentage. To investigate the 
principles of a riveted joint, it will be convenient to use 
a little algebraic notation. 

Let d indicate diameter of rivet ; s, section of sheet 
between rivets; p, pitch of rivets; t, thickness of sheet; 
c, per cent, of strength of riveted joint; m, tensile 
strength per square inch of sheet; n, shearing strength 
per square inch section of rivets in single shear: then 
.7851 d"2n = stm for equal strength of rivet and section. 
Let m divided by n equal r and 1 divided by .7854 
equal 1.273, then d^^ 1.273 srt, d = l 128-y'srt; and s = 
d^ -^ 1.273 rt; in all cases, s + d = pitch, or p. In this 
case, the strains on s and d are equal ; and the percent- 
age of strength of joint is s^p, or s-^-s+d = c. From 
this last equation, d = L273 rtc-f(l — c) ands= L273rtc2 
-^ (1 — c)-, in which c, the percentage of strength of 
joint, can be assumed. If the percentage of strength 
of joint is .56, d =z 1.62 rt, s = 2.06 rt, and p = 3.68 rt. 

Suppose the shearing strength rivet, 40000 ; tensile 
strength sheet, 50000 ; thickness sheet, j\ inch ; required 
percentage of single riveted joint, .70. — what shall be 
the diameter of rivets, lengths of sections between rivets, 



do THE ENGINEER S EPITOME. 

and pitch of rivets? In this case, s is 1.25, t is .3125, 
c is .70 ; and 1.273 x 1.25 x .3125 x .70 divided by 1 — 
.70 = 1.16 = diameter of rivet ; 1.273 x 1.25 x .3125 x .49 
divided by (1 — .70)2= 2.71 _ ^-,3^ section between rivets ; 
1.16+2.71 = 3.87 = pitch of rivets. If the pitch of the 
rivet is assumed, the diameter of the rivet must not be 
less than .845 -v'ptn, or the strength of section of rivet 
will be less than .56 ; and, if the pitch is taken less than 
3.68 rt, the section of sheet will be less than .56. 

With the same strength of rivet, sheet, and thickness 
as above, assume a pitch of 2| inches, then d = .845^ 
-^'2.25 x 3.125 X 1.25 = .792 diameter of rivets, s = 2.25 — 
.792 = 1.458 = section between rivets ; and, as this is 
greater than 2.06 rt, the section between rivets is greater 
than .56. In this case, the diameter of rivets could be 
increased and the strength of the joint made greater. 

The diameter of rivet to make the strongest joint, 

when the pitch is given, is 1.128 rt f y — — [--318 — •564 ) 

and the section between rivets is of equal strength. 

Using the elements as given in the last example, the 
diameter of rivet is .838 ; length of section between 
rivets, 1.412; and the percentage of joint is .628. Here, 
as is seen, the pitch is greater than 3.68 rt. Again, 
assume a pitch of 1:^ inches, which is less than 3.68 rt, 
the diameter of rivet for strongest joint is .579 ; length 
of section between joints, .671 ; percentage of strength, 
.537. If either the diameter of rivet or section of sheet 
between rivets is taken greater than 1.62 rt or 2.06 rt, the 
other remaining 1.62 rt or 2.06 rt, the percentage of joint 
will always be less than .56 per cent, of solid sheet ; and, 
if either is taken less than 1.62 rt or 2.06 rt, the strength 
of strongest joint will always be less than .56 of solid 
sheet. 



THE EXGINEER's EPITOME. 37 

Assume a diameter of rivet greater than 1.62 rt, — say, 
1.75 rt, — the section between rivets remaining 2.06 rt, 
the percentage of joint will be 2.06 divided by 2.06-1- 
1.7.5 = .5407. Assume the section of sheet between 
rivets greater than 2.06, — say, 2.50 rt, — the diameter 
remaining 1.62 rt, the rivets are weakest, and the per- 
centage of joint is 1.62 x 1.62 x .7834 divided by 1.82 = 
2.50 = .198 per cent, of solid sheet. 

Let grt indicate diameter of rivet, g being any num- 
ber greater or less than 1.62, then s = .7854 g-rt = section 
of equal strength ; and the percentage of strongest joint 
that can be made with that size of rivet is 1 divided 
, , 1.273 . 

byl+— 

Let xrt indicate section betw^een rivets, x being any 
number greater or less than 2.06, then d = 1.128 rt^^' x; 
and the percentage of strongest joint is 1 divided by 

^X 

Suppose the shearing strength of rivet to be 42000 ; 
tensile strength of sheet, 48000 ; thickness of sheet, ^ 
inch; diameter of rivet, | inch, — what is the percentage 
of strongest joint.? 

48000-1-42000= 1.143 

1.143 X .25 =.2857 = rt 

f = .625 

. 625^. 2857 = 2. 19 = g 

1.273^2 19 = .581 

I divided by 1 -]- .581 = .632 = percentage of strongest joint. 

2.19 X 2.19 X .7854 X .2857 = 1,075 = section between rivets. 

1.075 + .625 = 1.7 = pitch. 

Given the strength and thickness as above, and section 
between rivets | inch, — what is the percentage of strong- 
est joint? 



38 THE Engineer's epitome. 

I = .75 

.75 -f .2857 = 2.63 = x 

'^'2!63 = 1.62 

1.128 ^1.62 = .696 

1 divided by 1 + -696 = .589 = per cent, of joint. 

1.62 X .2857 X 1.128 = .522 = diameter of rivet. 

.75 + .522 = 1.273 = pitch. 

In the double riveted joint, using the same notation, 
we have 2 x .7854 d-n = stm, from which d = .797 -yJ srt 
and s=^1.57d^ divided by rt; or d = .637 rtc divided by 
(1 — c) and s = .637 rtc'^ divided by (1 — c)^, where c, 
the percentage of joint, is given. 

The strength of a double riveted joint is usually taken 
as .70 per cent, of solid sheet, at which percentage d = 
1.484 rt, s = 3.464 rt, and pitch = 4.948 rt. If the pitch 
IS given, the diameter of rivet to give a percentage of 
.70 of solid sheet is .667 a/ prt ; and, as in the single 
riveted joint, if pitch is taken less than 4.948 rt, the 
section of sheet between rivets will be less than .70 of 
solid sheet. 

If the pitch is given, the diametpr of rivet to make 

the strongest joint is .797 rt ( v/ — -1-.173 — .399 ). 

If the diameter is assumed, grt, s = 1.57 g'-^rt, and the 

637 
percentage of strongest joint is 1 divided by 1 + '—^ 

If the section between rivets is assumed, xrt, d = 
.797rt'V('x, and the percentage of strongest joint is 

-. n- .-. . 1 -. -"97 

1 divided by 1-j p^- 

^x 
As in the single riveted joint, if g or x is taken 
greater than 1.484 or 3.464, the other remaining 1.484 
or 3.464, the percentage of joint will be less than .70; 
and, if either is taken less than 1.484 or 3.464, the per- 
centage of strongest joint will be less than .70. 



THE engineer's EPITOME. 39 

In a triple riveted joint we have : 3 x .78-34 d-n = stm, 
from which d = .651 -yl srt and s = 2.356 d- divided by rt; 
or d = .424 rtc divided by (1 — c) and s = .424 rtc^ divided 
by(l-c)^. 

Taking the percentage of a triple riveted joint as .778, 
the diameter of rivets is very nearly the same as for 
double riveted joints, d being equal to 1.488 rt, s to 
5.208 rt, and pitch to 6.696 rt. 

If the pitch is given, the diameter of rivet to give 
a percentage of .778 of solid sheet is .575A(prt, and 
the diameter of rivet for strongest joint is .651 rt 



W 



^+.106— .325). If d equals grt, s = 2.356 g^rt. 



and percentage of strongest joint is 1 divided by 

.424 

1-j —. If s = xrt, d = .651 rt -yj x, and the percentage 

^ .651 

of strongest joint is 1-divided by 1-| — ^= 

\x 

In single welt butt joints the calculations are the 
same as for lap joints. In double welt butt joints the 
same calculations will apply, except that the rivets 
being in double shear, their strength is greatly increased, 
being generally considered as 75000 pounds per square 
inch of section for iron rivets. Hence n = 75000, and 
m divided by n or r is generally less than unity. 

If the diameters of rivets for all thicknesses of sheets 
are found by the foregoing rules, it will be seen that, 
as the thickness of sheet increases, the pitch of rivet to 
make a joint of .56 of solid sheet would soon be too 
great for boiler work. Consequently, the diameter of 
rivet and the pitch must be reduced, thus making a 
weaker joint. For this reason a double and triple 
riveted joint must be used, thus increasing the pitch 
and using more and smaller rivets. A very convenient 



40 THE engineer's epitome. 

and common diameter of rivets (using the rearost Jg to 
the result) is found by this formula: d= ■v' t-i-.25. For 
the section between rivets to give a per cent, of .56 of 
solid sheet, s = 1.273 d, or 1.273 ^t + . 25. The area of 
the rivet is .7854 (t+. 25). The section between rivets to 

make the strongest joint is - — '- — ^ ' ""' , and the percent- 

1.273 rt for 
age of strongest joint is 1 divided by 1 plus , — 

single rivet and 1 divided by 1 plus ', for double 

^ "^ Ajt+.25 

.424 rt 
rivet and 1 divided by 1 plus , for triple rivet. 

The diameter of rivet being given, the percentage of 

strongest joint is, for single rivet, ; for 

nd+ 1.273 mt 

double rivet, ; and for triple rivet, 

nd + .637 mt 
nd 

nd + .424mt' 

From all the foregoing we derive the following rules : 
For proper diameter of rivet, Rule : Add .25 to the 
thickness of sheet, and extract the square root of the 
sum. 

For area of rivet. Rule : Add .25 to the thickness of 
sheet, and multiply the sum by .7854. 

For section between rivets. Rule : Add .25 to the 
thickness of sheet. Multiply the sum by .7854 and the 
shearing strength per square inch section of rivet. 
Divide the product by the thickness of the sheet mul- 
tiplied by the tensile strength per square inch of section. 

To find the percentage of strength of strongest joint, 
Rule : Divide the diameter of rivet multiplied by its 



THE ENGINEER'S EPITOME. 41 

shearing strength per square inch of section by diameter 
multiplied by shearing strength, plus 1.273 times the 
thickness of sheet multiplied by tensile strength of sheet 
per square inch of section (for single riveted joints), plus 
.637 times thickness multiplied by strength (for double 
riveted joints), and by 1 plus .424 times thickness mul- 
tiplied by strength (for triple riveted joints). 

Example. — Shearing strength, 38000 ; tensile strength, 
52000; thickness, ^ or .5. What are the proper diameter 
of rivet and percentage of strongest joint triple rivet? 

.5 + .25 = .75 

V.75 = .86 (which call \^) = diameter of rivet. 

.875 X 38000 = 33250 

.424 X.5=-.212 

.212X52000=11024 

332.50+11024 = 44274 * 

33250 -7- 44274 = .751 = percentage of strongest joint. 

Diameters and percentages of strength of joint from 
above example with different thicknesses of sheet : — 



Thickness diameters. 


Single. 


Double. 


Triple 


1 

4- 


usei| 


) 




61 


76 


82 


i 


T6 


>per 


cent. 


50 


67 


78 


1 








39 


57 


66 



to find closely approximate percentage of 

Strongest Joint when Pitch of Rivets is given. 

— Rule: Multiply the thickness of sheet by its tensile 

strength per square inch of section. Divide the product 

by the shearing strength of rivets per square inch of 

section. Multiply the quotient by 220. Divide the 

product thus found by the pitch of rivets multiplied by 

number of rows plus three times the quotient found 

above. This last quotient is the approximate percentage 

of strongest joint. 

... 220 St 

Formula : 

np + 3rt 



42 



THE ENGINEER S EPITOME. 



The strain on the circumferential seams is but one- 
half the strain on the longitudinal seams. To prove 
this, suppose the pressure to be P pounds per square 
inch and the diameter D inches: then D"^ x .7851 x P = 
the force which the circumferential 
seam must hold. But there are 
D X 3.141 inches of seam to hold it, 
and the former expression divided 
by the latter equals the steam per 
square inch of seam. Now, the 
strain on the longitudinal seam 
equals ^ D x P, while the quotient 
above equals ^ D x P, or one-half 
the strain on the longitudinal seam. The strain on the 
headsr of boilers, however, is generally much less than 
this, as the areas of all the flues and tubes must be 
subtracted from the area of the head, and they amount 
in many cases to the area of the head. The working 
strain allowed on a boiler stay is 5000 pounds per square 





c<£. 



inch of section, and the unprotected parts of flat boiler 
heads must be protected by such a number of stays ' 
that the strain on each shall not be greater than 5000 
pounds per square inch of section smallest part of stay. 
In tubular boilers, the tubes, if properly placed, will 
support the part of the head they occupy ; and the cir- 



THE engineer's EPITOME. 43 

cumferential seam will support a rim about three inches 
wide all around the circumference, leaving the space to 
be stayed a segment of a circle comprising the space 
from about two inches above the tubes and three inches 
inside the circumference. Multiplying this space by the 
pressure and dividing the product by 5000 will give the 
number of stays of one square inch section each. The 
number of stays will also depend upon the thickness of 
the sheet, as they must be so near together that the sheet 
will not bulge between them. 

The approximate area of the segment of a circle is 
found by the following rule : Multiply the chord, ab, by 
the height, cd. Multiply this product by .685 plus .1 of 
the square of the height, cd, divided by the radius of 
the circle of which abd is a part. 

Formula: ch (.685 +1^)^ 
in which c indicates the chord, h the height, R the radius 
of the circle. 

A correct formula for finding the segment of a circle 
is as follows : Divide the height of segment by diameter 
of circle, and call the quotient N. Then 

Formula : (1.33123 — .831 N) — (.12 W) ^jW 

To FIND THE Area of the Segment of a Circle 
WHEN THE Part of Circumference included in the 
Segment can be measured. — Rule : From the meas- 
ured part of the circumference included in the segment 
subtract the chord. Multiply the remainder by the ra- 
dius of the circle. To the product add the product of 
the chord multiplied by the height of the segment. 
Divide the sum by 2. 

Formula: 5^^^?^=^^+^ 

9 

in which m indicates measured part of circumference. 



44 THE engineer's epitome. 

To FIND Distance between Centres of Stays. — 
Rule : Multiply the thickness by 190. Divide the prod- 
uct by the square root of the pressure. 

Formula : , - 

Example. — The sheet is \ inch thick, the pressure 90 
pounds. How far apart should be the centres of the 
stays ? 

I X 190 = 47.5 

-^90 = 9.486 

47.5 -i- 9.486 = 5.22 inches. Ansicer. 

The area of a stay equals the pressure multiplied by 
the area of the surface held by the stay divided by 5000. 

The diameter of a stay equals the square root of the 
pressure multiplied by the distance betvsreen stays and 
by .016. 

Formula : .016 s^/V 

What would be the diameter of the stay in the above 
example ? 



^/90 = 9 
9.486 X ? 


486 










.22 = 


= 49.516 








49.516 X 


.016 = 


= .79, diameter 


in smallest section. 


A 


nswer 



To FIND THE Thickness of the Sheet. — Kule : 
Multiply the distance between stays by the square root 
of pressure, and divide the product by 190 for iron or 
250 for steel. 

sa/p 
Formula • 



190 

Example. — The pressure is 120 pounds, distance be- 
tween stays 8 inches. What is the thickness of the iron. 



THE engineer's EPITOME. 45 

-^'120 = 10.9 

10.9 X 8 = 8.72 

8.72 -f 190 = .458 inch. Ansiver. 

To FIND THE Working Pressure for Stayed Sur- 
faces. — Rule: Multiply the square of the thickness of 
the sheet by 36000 for iron or by 60000 for steel, and 
divide the product by the square of the distance between 
stays. 

36000 t^ 
Jj ormuJa : ^ 

s- 

In these rules the stays are supposed to be spaced the 
same distance each way. 

Example. — The thickness of iron sheet is f^ inches, 
stays 7 inches apart. What is the working pressure ? 

jSg. == .312.5 expressed as a decimal. 

.3125-^= .0976 

.0976 X 36000 = 3513.6 

72 = 49 

3513.6 -f 49 = 71.7 pounds. Ansiver. 

To FIND Thickness of Circular Sheet. — Rule: 
Multiply the radius of the circle by the square root of 
the pressure, and divide the product by 71 for cast iron, 
or by 110 for wrought iron, or 111 for steel. 

Formula : — — -- — 

Example. — How thick should be the cast iron head of 
a boiler 36 inches in diameter, unstayed, to sustain a 
working pressure of 30 pounds per square inch ? 

-^.30 = 5.476 

The radius is 18 inches, or ^ the diameter. 

18 X 5.476 = 98.568 

98.568 -i- 71 = 1.388 inches. Answer. 



46 THE engineer's epitome. 

To FIND THE Thickness for Rectangular Sheets, 
Unstayed, Screwed at the Edges. — Rule: Divide 
the longest side by the shortest, and call the quotient N. 
Multiply the longest side by this quotient and by the 
square root of the pressure divided by 1 plus the quotient 
raised to the fourth power. Divide the product by 82 
for cast iron, 127 for wrought iron, and 165 for steel. 



]S[m X 
Formula : 



\/(n^0 



127 
in which m indicates the longest side. 

Example. — What should be the thickness of a wrought 
iron sheet screwed at the edges, unstayed, the length 
being 40 inches, width 30 inches, working pressure 20 
pounds per square inch ? 

40^30 = 1.33 = N 

1.33-^ = 3.1268 

1 +3.1268 = 4.1268 

20 -f_4.1268 = 4.84 

-^4.'84'= 2.2 

40 >^1.33 = .53.2 

.53.2 X 2.2 = 117.04 

117.04 -f 127 = .921 inches. Ansiver. 

To FIND THE Thickness of Square Plates, Un- 
stayed AND Screwed at the Edges. — Rule: Multi- 
ply the side of the square by the square root of the 
pressure. Divide the product by 115 for cast iron, 179 
for wrought iron, and 230 for steel. 



Formula : 



Q-yp 



179 
in which m indicates side of square. 

Example. — The sides of an unstayed square sheet of 
wrought iron are 35 inches, the working pressure is 20 
pounds. What is the thickness of the sheet ? 



THE engineer's EPITOME. 47 

-^20^ 4.47 

35 X4.47 = 156.45 

156.45 -f J79 = .874 inches. Ansive7\ 

To FIND THE Working Pressure of an Unstayed 
Circular Sheet, Screwed at the Edges, the Thick- 
ness BEING GIVEN. — Rule: Multiply the square of the 
thickness by 5000 for cast iron, 12000 for wrought iron, 
and 20000 for steel. Divide the product by the square 
of the radius of Jthe circle. 
12000 12 

Formula : — — 

K" 

Example. — A boiler 48 inches in diameter has an un- 
stayed cast iron head one inch in thickness. What is 
the working pressure ? 
r-=i 

1 X 5000 = 5000 

48 -r 2 = 24 = radius of circle. 

24- = 576 

5000 -7- 576 = 8.68 pounds. Ansiver. 

To FIND THE Working Pressure on an Unstayed 
Rectangular Sheet, the Thickness being given. — 
Rule : Multiply the square of the thickness by 6750 for 
cast iron, 16000 for wrought iron, and 27000 for steel, 
and by 1 plus the fourth power of the quotient found 
by dividing the longest by the shortest side. Divide the 
product by the product of the square of the longest side 
multiplied by the quotient. 

Formula: l^M^lIi±il3 
(mn)'2 

in which m indicates the longest side, and n the quotient 
of the longest side divided by the shortest. 

Example. — What is the working pressure on an un- 
stayed rectangular sheet of steel | inch thick, size 30 
by 60 inches V 



48 THE engineer's epitome. 



.875--^ = .7656 

6750 X .7656 = 20671.2 

60 ^ 30 = 2 

2*= 16 

l-fl6 = 17 

17 X 20671.2 = 351410.4 

2X 60= 120 

]20-2= 14400 

351410.4 -f 14400= 24.4 poimds. Ansiver. 

To FIND THE Working Pressure om an Unstayed 
Square Sheet, the Thickness being given. — Rule: 
Multiply the square of the thickness by 13000 for cast 
iron, 32000 for wrought iron, and 52000 for steel. Divide 
the product by the square of the side. 
32000 t2 

Formula : — 

in which m indicates the side of the sheet. 

Example. — The side of a cubical cast iron tank is 50 
inches, the thickness ^ an inch. What safe pressure will 
it stand ? , 

i = .5 

.5-2 = .25 

13000 X .25 = 3250 

50^ = 2500 

3250 ^- 2500 = ] .3 pounds. Answer. 

Steam domes built on the top of cylinder boilers 
should always be double riveted where they are attached 
to the boiler, and then heads stayed to the upright sheet 
of the dome, and not to the sheet of the boiler inside 
the dome, as the sheet of the boiler inside the dome is in 
equilibrium of pressure, and all the support it would 
give to the head would be that due to its stiffness. 
Where pipes are attached to a boiler, the boiler should 
be reinforced by a strengthening plate. Government 



TH'E engineer's EPITOME. 49 

inspectors require this plate for all sizes of pipes. Pipes 
of over two inches in diameter should be connected by a 
cast-iron flange riveted to the boiler. 

Following are short rules for finding the approximate 
working pressure of a cylinder boiler : — 

To FIND Approximate Working Pressure of 
Single Rivet Cylinder Boiler. — Rule : Di^nde 1000 
times the thickness of sheet, expressed as a decimal, by 
1 ^ times the diameter. 

To find Approximate Working Pressure of 
Double Rivet Cylinder Boiler. — Rule: Divide 
1000 times the thickness of sheet, expressed as a deci- 
mal, by the diameter; tensile strength, 52000; factor of 
safety, 6. 

General rule for finding working pressure of cylinder 
boilers : Multiply the lowest tensile strength of any sheet 
in the boiler by the thickness and by .56 for single rivet 
or by .70 for double rivet. Divide the product by ^ the 
diameter for bursting pressure. Divide this quotient by 
a factor of safety for working pressure. Factor of safety, 
from 4 to 8; generally, 5 for stationery boilers, 6 for 
government inspector, and | the hydrostatic test. 

Examples. — A single riveted cylinder boiler is 54 
inches in diameter ; thickness of sheets, ^ inch ; tensile 
strength, 52000 ; shearing strength, 40000 ; dome, 24 
inches in diameter ; height, 30 inches ; distance from top 
of tubes to shell, 22 inches. Find the diameter aud 
pitch of rivets, percentage of strength of seam, working 
pressure, and number and size of stays in dome and 
boiler head. 



50 THE engineer's EPITOME. 

2 X .25 = .50 

-W-SO = .707 = diameter of rivet. 
Call .707 \l, which is .6875 
..50 X .7854 = .3927 
f 2-Qflo = 13 ^ J. and rt = ..325 
.3927 -^ .325 = 1.208 = section between rivets. 
.6875+ 1.208= 1.895 

1.208 H- 1.895 = .637 = per cent strength of seam. 
i of 52000 = 13000 
.637 X 13000 = 8281 
8281 -f 27 = 306.7 = bursting pressure 
306.7 ^ 6 = 51.1 
24 — 6 = 18 

18^ = 324 = inches to be stayed. 
51.1 X 324= 16556 4 

16556.4 -f- 7 = 2365.2 = pounds held by one stay. 
5L5 = 7.14 



V' 



6.8 X 7.14 = 48.55 

.016 X 48.55 = .776 = diameter of stay. 

i of 190 = 47.5 

47.5 -f 7.15 = 6.6 = pitch of stays. 

6.6" = 43.56 = area held by one stay. 

324 -^ 43.56 = 7.4 = number of stays. 

(Use 7 instead of 7.4.) 

324 ^ 7 = 46 2 

-^46. 2 = 6.7 = pitch for seven stays. 

To FIND Segment of the Head. — Rule: The dis- 
tance from tubes to shell is 22 inches. Subtracting 3 
for circumferential seam and 2 for tubes, leaves 17 as 
height of segment to be stayed. In practice this height 
and corresponding chord will be found by direct measure- 
ment. In this case, it is very nearly 46 inches. 

To FIND the Chord when the Height is given. — 
Rule : Subtract the height from radius of circle. Sub- 
tract the square of the remainder from the square of the 
radius. Multiply the square root of the last remainder 
by 2. In practice the head sheets are made thicker than 



THE ENGINEER S EPITOME. 51 

the circumferential sheets, and are otherwise strength- 
ened by cast-iron flanges, to which the stays are fastened, 
thus considerably reducing the number of head stays, — 
ill this case, to 8 or 10. 

Many might put in smaller stays, and more of them ; 
but I would recommend that the pitch of stays be as 
large as the thickness of the sheet will allow, and that 
larger stays should be put in, as this leaves the inside of 
the boiler more open for the circulation of the water and 
for cleaning purposes. These stays in the fire-box, or 
where the sheets are parallel and but a small distance 
apart, are called " socket bolts," from the fact that they 
have around them, between the sheets, a piece of iron. 
They are usually screwed through both seats and riveted 
over the ends. The diameter of stays as given by those 
rules is for smallest section, and it is supposed that they 
are secured to the sheets in such a manner that the point 
of junction is as small as the smallest section of the stay. 

To find the collapsing pressure on riveted flues to 
correspond with Fairburn's Tables, multiply the square 
of the thickness of the sheet by 738000. Divide the 
product by the length in feet multiplied by the diameter 
in inches. 

738000 t^ 

Formula : -—- 

DL 

This rule, with a factor of safety of 8, compares very 

nearly with government inspectors' rule for safe working 

pressure, which is as follows : Multiply 89600 by the 

square of the thickness of the sheet. Divide the product 

by the length in feet multiplied by the diameter in 

inches, provided the length is not more than 8 feet and 

the diameter is more than 42 inches. If the length is 

more than 8 feet, it must be divided and strengthened 

by a flange with a wrought iron ring 2^ inches wide and 

not less than ^ inch thick, riveted between the flanges. 



52 THE KXGIXEEIl S EPITOME. 

Examples. — What pressure is allowed by inspectors' 
rule on a flue 30 inches in diameter, 21 feet long, and | 
inch thick ? 

In this case, the flue must have two flanges, dividing it 
into three parts, each seven feet long. 

I = .375 

.375- = .1406 

.1406X89600 = 12597.76 

7 X 20= 140 

12597.76 -^ 140 = 89.98 pounds. Answer. 

What are the collapsing and safe working pressures on 
a flue 40 inches in diameter, 8 feet long, and ^ inch 
thick ? 

738000 Xi= 184500 

40 X 8 = 320 

184500 ~ 320 = 576.5 = collapsiug pressure. 

576.5 -^ 8 = 72 = safe working pressure. 

The thickness of lap-welded tubes less than 10 inches 
in diameter is ^ig- of the square root of the diameter 
(J^a/D ). 225 times the thickness squared equals the 
diameter. A proper diameter of tube of any length is | 
the length of the tube in feet in inches. 

Examples. — What is the thickness of a tube 6 inches 
in diameter, and what is its proper length ? 

a/6^ 2.44 

jV of 2.44= .16 inch. 

The length in feet can be found by multiplying the 
diameter expressed in inches by 4. Hence 6 x 4 = 24 
feet, length. 

The thickness of a tube is .1 inch. What is its di- 
ameter and proper length ? 

.1-2 = .01 

.01 X 225 = 2.25 = diameter. 

2.25 X 4 = 9 feet = length. 



THE engineer's EPITOME. 53 

The ratio of heating surface to grate surface and of 
grate surface to tube areas varies somewhat in most in- 
stances, and a variation will be required for different 
kinds of fuel. But a good general rule for natural draft 
boilers is as follows : Make the grate area -^^ the heating 
surface, and the tube area 18 times the grate area in feet 
in inches. The stack should be no smaller in its small- 
est part than the area of all the flues or tubes. 

The area of the safety-valve should be ^^ of heating 
surface in square feet, taken in square inches, or one- 
half the grate area, taken in square inches. The heating 
surface comprises all the surface in which the flames and 
hot gapes are in direct connection with the water in the 
boiler, with but a thickness of metal between. Super- 
heating surfaces are those where the hot gapes have 
direct connection with the steam after it is formed, but 
with the thickness of sheet between. 

The heating surface should be so proportioned that the 
temperature of the gapes on leaving connection with the 
steam space in the boiler should not be more than one or 
two hundred degrees hotter than the temperature due to 
the steam. 

To FIND THE Area of Safety-valve in Square 
Inches to discharge all the Steam Equal to 
THE Maximum Evaporation, so that the Pressure 
shall not rise above the Absolute Pressure. — 
Rule : Divide the weight of water evaporated per hour 
by 52 times the absolute pressure. This rule is only 
correct when the gauge pressure is about 20 pounds. For 
all pressures above 20 pounds, the area found by this 
rule is a little too large. 

A partially correct rule for all gauge pressures is : Add 
1, jL of gauge pressure, and the square root of gauge 
pressure together. Multiply the sum of the three by 2^. 



54 THE engineer's epitome. 

To the product add the gauge pressure. Multiply this 
last sum by 48, and divide the evaporation in pounds i)er 
hour by the product. 

Example. — The gauge pressure is 75 pounds; evapo- 
ration, 3000 pounds per hour. What area of opening of 
valve will discharge the steam ? 

jV jofjS = 7.5 

-i/75 = 8.66 

1 + 7.5 + 8.66= 17.16 

2iX 17.16 = 42.9 

42.9 + 75 = 117.9 

48X117.9 = 5659.2 

3000 -f 5659 = .53 square inch. 

The Area of the Opening, Lift of Seat, and 
Bevel of Seat given, to find the Diameter of 
Valve. — Rule : Divide the given area by one-half the 
square root of bevel of seat, minus 1.2, and multiplied 
by lift of valve. Bevel of seat from 20 to 70 degrees. 

Formula : 
in which h indicates lift of valve, and s semi-angle of 
bevel. 

With an area of .332 square inch, a bevel of seat of 
45°, and a lift of .1 inch, what should be the diameter of 
valve ? 

^,'45 = 6.71 

i of 6.71=3.35 

3.35—1.2 = 2.15 

,1X2.15 = .215 

.332 -— .215 = 1.54 inches. Answer. 

If the valve has a seat of 90, or one which is flat, divide 
the area by 3.14 times the lift, in finding the diameter. 

In Lever-weighted Safety-valves, the Area, 
Pressure, and Weight being given, to find the 
Distance of Weight from Fulcrum, and. Distance 
FROM Fulcrum given, to find Weight. — Rule : 



THE engineer's EPITOME. 55 

Find weight of valve, of lever, and the distance of its 
centre of gravity from the fulcrum. Multiply the area 
of the valve by the pressure and by the distance from 
the fulcrum to the valve. Subtract from the product the 
weight of the valve multiplied by the distance from 
the fulcrum to the valve and the weight of the lever mul- 
tiplied by the distance of its centre of gravity from 
the fulcrum, and divide the remainder by the length of 
the lever from the fulcrum for the weight, and divide the 
remainder by the weight for the distance of the weight 
from the fulcrum. 

Let a indicate area valve. 

Let o indicate gauge pressure. 

Let f indicate distance from fulcrum to valve. 

Let h indicate weight of valve. 

Let r indicate distance from fulcrum to centre of 
gravity of lever. 

Let g indicate weight of lever. 

Let 1 indicate length of lever. 

Let w indicate weight. 

By transposing these formulas, any one of the amounts 
can easily be found. 

Examples. — The area of a safety-valve is 12 square 
inches, gauge pressure 70 pounds, weight of valve 15 
pounds, distance from fulcrum to valve 4 inches', weight 
of lever 20 pounds, length of lever from fulcrum to 
weight 36 inches, from fulcrum to centre of gravity of 
lever 16 inches. AVhat is the weight ? 

70 X 12 = 840 

4 X 840 = 3360. 

4X 15 = 60 

3360 — 60 = 3300 

16X20 = 320 

3300 — 320 = 2980 

2980 -^ 36 = 82.77 pounds. Answer. 



56 THE engineer's epitome. 

The area of valve is 9 square inches, pressure 80 
pounds, valve weight 10 pounds, distance from fulcram 
to valve 3 inches, weight of lever 14 pounds, distance 
from fulcrum to centre of gravity of lever 12 inches. 
At what distance from the fulcrum must a weight of 
70 pounds be applied ? 

80 X 9 -= 720 

3 X 720 = 2160 

3 X 10 = 30 

2160 — 30 = 2130 

12 X 14 = 168 

2130—168 = 1962 

1926 -f- 70 = 28.02 inches. Answer. 

A stop-valve should never be placed between the boiler 
and safety-valve, and their being so placed has been the 
cause of many disastrous explosions. 

In a battery of boilers, each boiler should have its own 
safety-valve, and should be connected to the boiler inde- 
pendently of any other connection. 

The side of a spiral spring of square steel on a spring- 
loaded safety-valve equals .018 times the square root of 
the pressure on the valve, or .021 times the square root of 
the pressure on the valve for a spring of round steel. 
The mean diameter of the coil of the spring equals 11000 
times the cube of the side of square steel divided by the 
pressure on the valve, or 8000 times the cube of the 
diameter of round steel divided by the pressure on the 
valve. The pitch of distance from the centre of one coil 
to another is twice the side or diameter of the steel. 
The distance or amount that the spring must be com- 
pressed in order to balance a given weight equals the 
weight nmltiplied by the cube of the mean diameter of 
the coil, multiplied by the number of coils, divided by 
the fourth power of the side of square steel, multiplied 
by 195000, or divided by the fourth power of the diameter 
of round steel, multiplied by 150000. 



THE engineer's EPITOME. 57 

Formulas : w indicates total weight on valve ; c, diam- 
eter of coil ; n, number of coils ; s, side of square steel ; 
d, diameter of round steel. 

.OlS-Ww = side of square steel. 
.021-v,'\v = diameter of round steel. 

11000 s^ 

= mean diameter of coil, square steel. 

8000 d^ = mean diameter of coil, round steel. 



2 s or 2 d = pitch 
wc^n 



195000 s^ 
wc'% 



= amount of compression to balance w, square steel. 

,,_: amount of compression to balance w, round steel. 
150000 d-^ ^ 

Example. — The area of safety-valve is 9 square inches, 

gauge pressure 120 pounds, number of coils 10. What 

is the side of the square steel of which the spring should 

be made ? What is the outside diameter of the coil, and 

how much must spring be compressed not to blow off 

until the gauge pressure is reached? 

120X9 = 10S0 = w 

-^1080 = 32.2 

.018 X 32.2 = .5796 = s (which call .6) 

.6-3 =.216 

.216 X 11000 = 2376 

2376 -M080-= 2.2 = c 

2.2 + .6 = 2.8 = outside diameter. 

2,23 = 10.648 

10.648 X 1080=11499.84 

10 X 11499.84= 114998.4 

.64 = .1296 

.1296 X 195000 = 25272 

114998.4 -r 25272 = 4.55 = compression of spring. 

The actual volume of steam in cubic feet that will 
escape from an orifice one square inch in area, in one 
minute, equals .26 times the theoretical velocity of the 
steam in feet per second. 



V 



1733472.14 = 1316.6 



58 THE engineer's epitome. 

The volume of steam in cubic feet that will escape 
from an orifice equals the theoretical velocity of the i 
steam in feet per. second, multiplied by the area of the ] 
orifice, multiplied by the time in minutes, and by from 
.42 to .26, depending upon the form of the orifice. The 
multiplier .26 gives the flow through a thin sheet which 
is I the theoretical flow, and the number used in practice. 

The weight of the steam is found by multiplying the 
number of cubic feet by the weight of one cubic foot, at 
the density due to the pressure. 

Example. — How many cubic feet of steam at 100 
pounds' gauge will flow into steam at 50 pounds' gauge 
in twenty minutes through an orifice of 2 square inches, 
and what is the weight of the steam? 

100 — 50 = 50 = difference in gauge pressures. 

50 X 666000 = 33300000 

100 X .163 = 16.3 

2.91 + 16.3 = 19.21 

33300000 -i- 19.21 =1733472.14 



1316.6 X 2 = 2633.2 

2633.2 X 20 = 52664 

.26 X 52664 = 13692.64 = cubic feet discharged in 20 minutes. 

-JlOO=10 

81 + 10 = 91 

.208 X 100 = 20.8 

20.8 + 3.125 = 23.925 

23.925 ^ 91 = .2629 = weight of 1 cubic foot. 

.2629 X 13692.64 = 3599.795 = weight discharged in 20 minutes. 



To FIND THE Average Theoretical Velocity at 
WHICH Steam will discharge into the Atmosphere 
until All is discharged, no Steam being formed 
AT the Time. — Rule: Divide 12 by the gauge pressure. 
Add .21 to the quotient. Divide 666000 by the sum, and 
extract the square root of the last quotient. 



THE ENGINEER S EPITOME. 59 

To FIND THE Time in which the Steam in a Boiler 

CAN be blown into THE AlR, NO StEAM BEING FORMED 

AT THE Time. — Rule: Divide 4 times the number of 
cubic feet of steam in the boiler by the average velocity, 
multiplied by the area of the orifice, or safety-valve. The 
quotient equals the sum in minutes. 

Example. — The steam space ia a boiler is 150 cubic 
feet, the gauge pressure is 70 pounds, the area of the 
valve is 4 square inches. In what time will the steam 
be discharged into the air? 

12^ 70 = . 17 

.17 + .21 =..38 

666000 ^ ..38 = 1752631. .57 



-Wl752631.5 = 1323.8 = average velocity. 

150X4 = 600 

1323.8 X 4 = 5295.2 

600 -f- 5295.2 = .113 minute. ^?2sit'er. 

To find the Time in which the Pressure will 

RISE FROM G TO G (G AND G ARE THE GaUGE PRESS- 
URES). — 966 multiplied by the pounds of water evapo- 
rated per pound of coal per hour equals the utilized 
thermal units of heat per pound of coal. Then multiply 
840 times the weight of water in the boiler in pounds 
by the difference of the square roots of the pressures. 
Divide the product by the utilized thermal units in 1 
pound of coal multiplied by the pounds of coal burned 
per hour. The quotient will be the time in minutes. 

840 w(V^^/^) 

Formula: ^-^ ^ 

tc 

in which w indicates weight of water ; t, thermal units 

of heat ; c, coal burned per hour. 

Example. — If 1 pound of coal evaporates 11 pounds of 

water from and at 212°, in what time will the gauge 

pressure rise from 20 to 75 pounds, the weight of water 



60 THE engineer's EPITOME. 

in the boiler being 15000 and rate of combustion being- 
SCO pounds of coal per hour ? 

966 X 11 = 10626 = utilized thermal units of heat. 

^20^ 4.47 

^75"= 8.66 

8.66 — 4.47 = 4.19 

15000 X 840 = 12600000 

12600000 X 4.19 = .52794000 

10626 X 300 = 3187800 

52794000 -!- 3187800 = 16.56 minutes. Answer. 

The H. P. of a boiler at the present time is generally- 
considered as the evaporation of 30 pounds of water from 
a temperature of 70° and a pressure of 100 pounds gauge, 
or an evaporation of 35f pounds from and at 212°. From 
12 to 15 square feet of heating surface are allowed in 
natural draft boilers to one H. P. The evaporation will 
probably average from 1 to 1 ^ gallons of water for every 
pound of coal burned, or from ^ to ^ of a cubic foot. 

To FIND THE Relative Evaporation from and 
AT 212°.— Rule: To 1 add .00104 times the difference 
between the temperature of the feed water and 212°, 
and .004 times the square root of the gauge pressure. 

Formula: 1 + .00101 (212 — f) + .004^^"" 
in which f indicates temperature of feed water. 

Example. — A boiler evaporates 9 5 pounds of water 
with 1 pound of coal when the temperature of the feed 
is 60° and the gauge pressure is 72 pounds. How much 
water will 1 pound of coal evaporate when the tempera- 
ture of the feed is 212° and the gauge pressure is 0? 

212°— 60 = 152 

152 X .00104 = .15808 

-^72^ 8.49 

8.49 X .004 = .03396 

1 + .15808= 1.15808 

1.15808+ .0.3396 = 1.192 

1.192 X 9.5 = 11,324 pounds. Answer. 



THE engineer's EPITOME. 61 



TESTS FOR MOISTURE IN STEAM- 
BARREL CALORIMETER. 



Set a barrel or small pail on a scale. Carefully weigh 
into it a definite amount of water. Connect a small 
steam-pipe with the pipe or space from which the steam 
to be tested is taken, in such a manner as to secure an 
average quality of the steam. Plug the end that goes 
into the pail or barrel, and drill a number of holes around 
it, so that the steam blown into the water will be equally 
distributed in the water. For a valve for shutting off 
the steam from the barrel, use a waste water cock, so 
that, when the steam is shut off, the waste water will 
open on the side next the barrel. This lets out the steam, 
and prevents a vacuum being formed in the lower part 
of the pipe after the steam is shut off, and draws a 
part of the water up the pipe. Put the valve close to the 
barrel, so that it will be convenient to get at. Note care- 
fully the temperature of the water before the steam is 
turned on, the temperature when the steam is turned off, 
and the increase of the weight of water in the barrel. 
These observations should be made as accurately as pos- 
sible. 

For finding the Percentage of Dry Steam, use 
the following rule : Divide the total weight of water 
after steam is blown in by the increase in weight of 
water. Multiply the quotient by the increase in temper- 
ature of the water. To the product add the temperature 



62 THE engixp:er's epitome. 

of the water before the steam was blown in. From this 
sum subtract the temperature of the steam due to the 
pressure. Divide the remainder by the latent heat of 
the steam due to the pressure. The quotient equals the 
percentage of dry steam. The greatest error likely to 
occur in making these tests, is that due to the rise in 
temperature of the water. By making the quantity of 
the water small and the increase in the temperature con- 
siderable, this error is reduced to a minimum. 

A good way is to weigh accurately 9 pounds of water 
into a pail. Set the weight to balance 10 pounds. Note 
the temperature of the water. Turn on the steam, and 
the instant it balances 10 pounds shut off steam and 
note the temperature. Note the steam pressure while 
the steam is turned on. Then, to calculate the percentage 
of water in the steam, use the following rule : From 1590 
subtract 10 times the observed rise in the temperature 
of the water and the temperature of the water before 
steam was admitted. Divide the remainder by 977 
minus 10 times the square root of the gauge pressure. 
From the quotient subtract .425. The remainder equals 
the jDercentage of water in the steam. If the remainder 
is less than 0, the steam is superheated ; if 1, no steam, 
but all water ; if more than 1, water at a temperature 
less than that due to pressure. The pipe and the vessel 
in which the trial is made should be protected to pre- 
vent chill and radiation ; and the pipe should be blown 
through to remove any water that might be in it, and to 
heat it before each trial. 

. (1590 — RIO) — t .^. 
formula: ^ / — .42o 

977-10^/g 

in which K indicates observed rise of temperature in 
water, t temperature of water before steam is admitted. ' 
The rise in temperature of the water due to the steam 



THE engineer's epitome. 63 

is the temperature of the water multiplied by its weight, 
plus the temperature of the steam multiplied by its 
weight, divided by the sum of the weights. From the 
quotient subtract the temperature of the water. 

The rise in the temperature of the water due to the 
water in the steam is the temperature of the water mul- 
tiplied by its weight, plus the temperature of the water 
in the steam multiplied by its weight, divided by the 
sum of the weights. From the quotient subtract the 
temperature of the water. 

Examples. — A vessel containing 100 pounds of water 
at a temperature of 65° has steam blown in until the 
weight is 105 pounds and temperature 95°. The gauge 
pressure is 64. What is the per cent, of dry steam ? 

105 — 100=5 
105-^5 = 21 
95 — 65 = 30 
21 X 30 = 630 
630 + 65 = 695 

^6r= 8 

14X8 = 112 

199+112 = 311= temperature of steam. 
695 — 311 =384 
6^=8 



V' 



10 X 8 = 80 

977 — 80 = 897 = latent heat. 

384 ^ 897 = .428 per cent. Answer. 

9 pounds of water at 70° are increased to 10 pounds 
at 180°. The gauge pressure is 81. What is the per 
cent, of dry steam ? 

10 — 9 = 1 

10-^ 1 = 10 

180 — 70 = 110 = observed rise. 

110X10=1100 

1590—1100 = 490 

490 — 70=420 



64 THE engineer's epitome. 

10X9 = 90 

977 — 90 = 887 = latent heat. 

420 ^ 887 = .473 

.473 — .425 = .048 per cent, of water. 

1 — .048 = .952 per cent, dry steam. Ansiver. 

What should be the rise in temperature of 50 pounds 
of water at a temperature of 40°, into which 6 pounds of 
dry steam are blown, at 50 pounds' gauge pressure ? 

^'50^7.07 

41 X 7.07 = 30.631, or 30.64 

1173.5 + 30.64 = 1204.14 = temperature of 1 pound of steam. 

6 X 1204.14 = 7224.84 = total heat of steam. 

50 X 40 = 2000 

7224.84 -1- 2000 = 9224.84 

50+6 = 56 

9224.84 -i- 56 = 164.7 

164.7 — 40 = 124.7 = rise in temperature. Answer. 

Fifteen pounds of water at temperature due to 25 
pounds gauge pressure are blown into 40 pounds of water 
at 32°. What is the resultant temperature of the water? 

^'25~=.5 

14 X 5 = 70 

199 + 70 = 269 = temperature of water. 
32 X 40 = 1280 
269 X 15=4035 
1280 + 4035 = 5315 

15 + 40 = 55 
5315 -f 55 = 96.6° Answer. 



THE engineer's EPITOME. 65 



CHIMNEYS. 



The duty of a chimney is to create a draft through the 
fuel in the furnace sufficient to maintain combustion and 
carry off the gases formed by the combustion. In the ab- 
sence of any apparatus for ascertaining the temperature 
in the flue or chimney, it can be approximately deter- 
mined by the height at which the draft will support a 
column of water, in the following manner : Connect a 
small pipe to the flue or chimney in some convenient 
place. To the lower end of the pipe attach a piece of 
water glass gauge. Make the connections tight. Make 
a zero mark on the glass, and divide it into tenths of an 
inch from the zero mark upward. A better way would 
be to attach a small scale graduated to one one-hundredth 
of an inch. If the pipe is to remain permanently, place 
a valve in it above the glass. Let the glass dip perpen- 
dicularly in a vessel of water, so that the zero mark will 
come exactly even with the surface of the water, and 
note accurately the height which the water in the glass 
rises above the zero line. Then, 

To FIND THE Temperature in the Flue, use the 
following rule : Divide the height of the water in deci- 
mals of an inch by .000257 multiplied by the height of 
the chimney. To the square of the quotient add the ex- 
ternal temperature. The sum equals the temperature in 
the flue. 



66 THE engineer's epitome. 

To FIND THE Theoretical Velocity at which 
External Air will flow into a Furnace. — Rule : 
To 461 add the temperature of the external air. Divide 
the sum by 461 plus the temperature of the internal air. 
Subtract the quotient from 1. Multiply the remainder 
by 64 times the height of the chimney. Extract the 
square root of the product. 

Formula: i / 64 h fl — ^^^-t^\ 
V V 461+t/ 

in which e indicates external temperature, t internal 

temperature, and h height of chimney. 

Example. — The height at which the draft in a chimney 

will support a column of water is .32, the chimney is 81 

feet high, the external atmosphere is 70°. What is the 

velocity iu feet per second at which the external air will 

flow into the furnace ? 

.000257 X 81 = .0208 

.32^ .0208= 15.384 

15.384-^ = 236,667 

236.667 + 70 = 306.667 = internal temperature. 

461+70 = 531 

461 +306.667 = 767.667 

f31 -^ 767.667 = .691 

1_. 691 = .309 

81 X 64= 5184 

.309 X 5184 = 1601.856 

-\/l601.856 = 40 Answer. 

To FIND THE Velocity at which the Internal. 
Gases will flow into the Air. — Rule: To 461 add 
the internal temperature. Divide the sum by 461 plus 
the external temperature. Subtract 1 from the quotient. 
Multiply the remainder by 64 times the height of the 
chimney. Extract the square root of the product. 



Formula: 4 / 64 h (^^?1±^ _ 
V M61 + e 



THE engineer's EPITOME. 67 

Exam2-)le. — The height of water column is .62 inch, 
height of chimney 100 feet, external temperature 45°. 
What is the velocity of internal gases? 

.000257 X 100 = .0257 

.62 ^ .0257 = 24.1 

24.1^ = 580.81 

580.81 -j- 45 = 625.81 = internal temperature. 

461 +625.81 = 1086.81 

461 + 45 = 506 

1086.81 ^ 506 = 2.14 

2.14— 1 = 1.14 

100 X 64 = 6400 

1.14 X 6400 = 7296 

-1/7296 = 85.41 Ansiver. 

The relative volume of the external and internal air 
is found thus : Divide 461 plus the internal temperature 
by 461 plus the external temperature. Now, as the pro- 
duct of the relative volume multiplied by the flow of ex- 
ternal air increases faster than the flow of internal gases 
into the air, there will be a temperature in the chimney 
at which the velocity of flow of external air into the fur- 
nace will increase; and, if the temperature in the chim- 
ney is increased beyond this, the flow of air to the 
furnace will decrease, as the flow of air into the furnace 
can only equal the flow of the hot gases into the air 
divided by the relative volume of the internal gases and 
external air. The actual velocity of air into the furnace 
will vary with different kinds of fuel and different thick- 
nesses of fuel on the grate, and can only be determined 
by experiment. From ^to ^ the theoretical velocity will 
be about the velocity in practice into the furnace, and, 
as at a relative volume of from 2 to 3, the maximum flow 
of air to the furnace is obtained, it will be seen that the 
best effects are produced when the temperature of the 
flue is such as to produce this relative volume. The flue 



68 THE ENGINEER S EPITOME. 

temperature to produce this result is from 500 to 600 



To FIND THE Amount of Coal burned ter Square 
Foot of Chimney per Hour in a Well-proportioned 
Boiler and Chimney. — Rule: From the square root of 
height of chimney subtract .5. Multiply the remaiader 
by 16. 

To FIND the Number of Pounds burned per 
Square Foot of Grate. — Rule : Multiply th« above 
result by the area of chimney in square feet. Divide the 
product by area of grate in square feet. 

About 250 cubic feet of air will be drawn into the 
furnace for every pound of coal burned, or about 18 
pounds at a temperature of 60°. 

The effect of a horizontal flue leading from the boiler 
to the chimney is to reduce the H. P. of the chimney 
about .14 per cent, for every time he length of the flue 
above 100 feet is doubled. For flues less than 100 feet, 
the loss is a like proportional part of .14 per cent. 

To FIND THE Loss FROM LONG HORIZONTAL FlUES. — 

Rule : Divide the length of the flue by 2. Divide this 
quotient by 2, and continue the division until the last 
quotient is between 100 and 200, unless the last quotient 
is 100. Add 1 to the number of times 2 was taken as a 
divisor. Multiply the sum by .14. Multiply the amount 
by which the last quotient was greater than 100 by .14. 
Divide the product by 100. Add the quotient to the 
product above. This is the percentage of loss, and must 
be subtracted from the x^ower of the chimney without 
the flue. 

Examples. — The area of a chimney is 5 square feet, 
the height is 85 feet. How many pounds of coal will be 
burned per hour? 



THE engineer's EPITOME. 69 

-y/85^ 9.219 

9.219 — .5 = 8.719 

8.719 X 16 = 139.504 

139.504 X 5 = 697.52 = pounds per hour. Answer. 

500 H. P. are required, and the flue to the chimney is 
1200 feet long.. What will be the percentage of loss 
from the horizontal flue? 

1200 -^ 2 = 600 

600 ^ 2 = 300 

300 4- 2 = 150 

1+3=4 

4 X. 14 = .56 

50 X. 14 = 7 

7-^100 =.07 

.56 + .07 = .63 = percentage of loss from horizontal flue. 

Answer. 
Explanation. — 2 is used as a divisor three times. Add 
1 to 3, and multiply by .14. The last quotient is 50 
greater than 100. Then multiply 50 by .14, and divide 
the product by 100. Add the quotient to .56, making 
.63, which is the percentage of loss from the horizontal 
flue. 

To FIND THE H. P. OF A Chimney. — Rule : Multiply 
the area in square inches by the square root of the height 
in feet. Divide the product by 84 minus twice the 
square root of the height. 



Formula : 



a^^/h 



84 — 2^'h 
Example. — The height of a chimney is 196 feet, the 
area is 40 square feet. What is the H. P.? 
40 X 144 = 5760 square inches. 
-WT96 = 14 
5760 X 14 = 80640 
14X2=28 
84 — 28=56 
80640 -!- 56 = 1440 = H. P. Ansiuer. 



70 THE engineer's EPITOME. 

To FIND THE Area. — Rule : Multiply the H. P. by 84 
minus twice the square root of the height. Divide the 
product by the square root of the height. 

Fo^ula: H- F- (84 - 2 yiT ) 

Examjjle. — The PI. P. is 60, the height is 70 feet. 
What is the side of the square chimney? ^ 



V' 



70 = 8.366 



8.366 X 2= 16.732 
84 — 16.732 = 67.268 
60 X 67.268 = 4036.08 
4036 08 -f 8.366 = 482.4 = area. 
J482.4 =21.96 = side of square. Answer. 



^4: 



To FIND THE Height of Chimney, the H. P. and 
Area in Square Inches being known. — Rule : Divide 
84 times the H. P. by the area in square inches plus 
twice the H. P. Square the quotient. 
, / 84H.P. \2 

^"^"^"'"^ (a + 21LP.) 

Example. — The internal diameter of a round chimney 
is 6 feet, the H. P. is 800. What should be the height 
of the chimney? 

62 = 36 

36 X .7854 == 28.27 square feet. 

28.27 X 144 = 4070.88 = area in square inches. 

84 X 800 = 67200 

800X2= 1600 

4070.88 + 1600 = 5670.88 

67200-^5670.88=11.84 

11.842 = 140.18 feet. Answer. 

The term " horse power " is assumed to be a weight 
or force of 33000 pounds raised or moved through the 
space of one foot in one minute of time. The jproduct 



THE engineer's EPITOME. 71 

of the three factors, — force, distance in feet, and time in 
minutes, — either of which may be a variable quantity, 
divided by 33000, gives the H. P. multiplied by the time 
in minutes. The H. P. of an engine is the force exerted 
on the piston by the pressure of the steam multiplied by 
the distance the piston moves in feet in one minute, 
divided by 33000. The distance the piston moves in feet 
in one minute is twice the stroke of the engine in feet 
multiplied by the number of revolutions of the crank 
shaft. 

The force on the piston equals the square of the di- 
ameter of the cylinder multiplied by .7854 and by the 
average effective pressure in the cylinder for the whole 
stroke. The effective pressure is usually found by the 
aid of an "indicator card." The usual way to find the 
average pressure from an "indicator card" is to divide 
the length of the card into ten equal parts. Divide the 
sum of the lengths measured in the centre of these spaces 
inside the card by 10. Multiply the quotient by the 
power or scale of the spring for the effective pressure- 

A good way to measure a card in the absence of a 
special apparatus is to fasten the card on a board or 
desk, so that the atmospheric line is parallel with the 
edge of the board. Draw lines perpendicular to the at- 
mospheric line through the card, — draw fifteen or twenty 
lines, the more the better, — spacing them as nearly alike 
as can be done by the eye. Add to ^ the sum of the ex- 
treme lines, the sum of the intermediate lines. Divide 
the amount by the number of lines less 1. Multiply the 
quotient by the scale of the spring. 

The pressure at any part of the stroke, by Mario tte's 
law for the expansion of gases, is equal to the initial 
pressure multiplied by the part of stroke performed be- 
fore cut-off plus clearance, divided by the part of stroke 



72 THE engineer's epitome. 

at which the pressure is required plus clearance- For 
example : If the stroke is 40 inches, the clearance .5 
inch, and the cut-off takes place when the piston has 
moved 10 inches, what is the pressure on the piston 
when it has moved 30 inches, the initial pressure being 
80 pounds ? The cut-off, 10, plus the clearance, .5, 
divided by the part of stroke moved, 30, plus the clear- 
ance, .5, equals .344. This multiplied by the initial 
pressure, 30, equals 27.52 pounds' pressure on the piston 
when it has moved f of its stroke. 

What would be the terminal pressure in this case ? 
The terminal pressure is the pressure at the end of the 
stroke due to expansion. In this case the terminal press- 
ure would be 10.5 divided by 40.5 multiplied by 80, 
which equals 20.72 pounds. The terminal pressure is 
also found by dividing the initial pressure by the ratio 
of expansion, and is the pressure used in determining 
the volume and weight of steam used by the engine. 

To FIND the Ratio of Expansion. — Rule : Divide 
the whole stroke plus the clearance by the cut-off plus 
the clearance. 

To FIND THE Part of Stroke in Expansion. — 
Rule : Divide the stroke minus the cut-off by the stroke 
plus the clearance, or divide 1 by the ratio of expansion 
and subtract the quotient from 1. This part of stroke 
in expansion is the amount to be used in finding the 
average pressure by calculation in the following rules. 
In calculating average and terminal pressures, the abso- 
lute pressures must in all cases be used. 

To find THE Average Pressure, Cut-off Less 
THAN ^L. Stroke. — Rule : From 1 subtract the square 
of the expansion divided by 1.03 plus 1.8 times 1 minus 



THE engineer's EPITOME. 73 

the expansion. Multiply the remainder by the initial 
pressure. 

Formula : p ( 1 , ) 

^ 1.03 + (1.8 X 1 — E)'-^ 

Examjjle. — The stroke of engine is 48 inches ; clear- 
ance, A inch ; cut-off, 2 inches ; initial pressure, 90 
pounds absolute ; back pressure, 2 pounds absolute. 
What is the effective pressure on the piston for the 
stroke ? 

48 — 2 = 46 

48 + .4 = 48.4 

46 -r 48.4 = .95 = expansion. 

.952= .9025 

1— .95 = .05 

1.8 X .05 = .09 

1.03 + .09 = 1.12 

.9025 -^ 1.12 = .8058 

1 — . 8058 = .1942 

.1942X90=17.478 

17.478 — 2 = 15.478 = effective pressure. Aiiswer. 

To FIND THE Average Pressure when the Cut-off 
IS Greater than Jq the Stroke. — Rule : From 1 
subtract the square of the expansion divided by 1.84 
minus .7 of the sum of the square of the expansion 
and -^-Q of the expansion. Multiply the remainder by 
the initial pressure. 

W 

Formula : 1 

1.84 — .7(E2 + .1E) 

Example. — The stroke of the engine is 60 inches ; 
cut-off at 15 inches ; clearance, .6 inch ; initial pressure, 
60 pounds' gauge ; back pressure, 1.5 pounds. What is 
the effective pressure on the piston for the stroke ? 

60 — 15 = 45 

60 + .6 = 60.6 

45 ^ 60.6 = .74 

.74^= .5476 

.074 +.5476 = .6216 



74 THE engineer's epitome. 

.7 X .6216 = .4351 

1.84 — .43.51 = 1.4049 

.5476^ 1.4049 = .389 

1_, 389 = .611 

75 X.611 =45.82 

45.82 — 16.5 = 29.32 = effective pressure. Answer. 



In calculating average pressures, the absolute pressure 
must always be used, and the back pressure calculated 
from 2000. 

To FIND THE H. P. OF AN Engine. — Rule: Multiply 
the square of the diameter in inches by the stroke in 
inches, by the number of revolutions per minute, and by 
4. Cut off six figures from the right. The result is the 
engine constant, or H. P. per 1 pound effective pressure. 
Multiply the result by effective pressure, for H. P. due to 
pressure. 

Formula : D- x S x R x .00000397 

If the piston-rod is taken into account, the formula 
will be as follows : — 



(--f) 



X S X R X .00000397 



The rule gives f H. P. too much in 100 H. P., but the 
formulas are correct. 

Example. — The cylinder of a non-condensing engine is 
30 inches in diameter; stroke, 60 inches; gauge pressure, 
85 ; back pressure, 2 pounds ; cut-off at 16 inches ; revo- 
lutions, 60 ; clearance, A inch of stroke. What are the 



THE ENGINEER S EPITOME. 



75 



calculated H. P., the pounds of water per H. P. per hour, 
and the pounds of coal per H, P. per hour.- if one pound 
of coal evaporates 9 pounds of water ? 

In finding the average pressure, (stroke + 1 clearance) — 
(distance steam follo^ys -f- 1 clearance) = ratio of expansion. 



16.4)60.400(3. 


68 ratio of exp. 


492 










1120 








.634 


984 








83 


1360 








1902 


1312 








5072 
52.622 av, press. 


lbs. 










85 gauge press. 




30' 


dia. 


piston = 706.86 sq. in. area. 


2 back press. 








600 ft. per mi 


83 initial press. 








33000)424116.00 

12.8 H. P. constant. 
52.6 av. press. 


60 in. stroke. 






673.28 H. P. 



120 in. to 1 rev. 
60 rev. per min. 
12 )7200 

600 ft. piston travel per min. 

706.8 sq. in. in piston. 
16.4 



11591.52 volume in cu. in. in cylinder at 16'' cut-off, 
270 cu. in. of steam from 1 of water at 85 lbs. press. 

270)11591.52 

42.9 cu. in. of water used m 1 stroke. 
2 

85.8 cu. in. of water used in 1 rev, 

80 

5148.0 cu, in, of water used in 1 min. 
60 
308880.0 cu. in. of water used in 1 hour. 



76 THE engineer's epitome. 

1728)308880.0(178.7 cu. ft. of water used in 1 hour. 
1728 
13608 
12096 
15120 
13824 
12960 
12096 

178.7 
62. .5 lbs. in 1 cu. foot. 
11168.75 lbs. of water used in 1 hour. 

H. P. 

673.28)11168.75 

16.5 lbs. of water per H. P. per hour. 

9)16.5(1.8 lbs. of coal per H. P. per hour. 
9 
"75 



In compound engines the total ratio of expansion in 
either cylinder is equal to the square of its diameter 
multiplied by its stroke plus its clearance, divided by the 
square of the diameter of the high pressure cylinder 
multiplied by its cut-off plus its clearance. The actual 
ratio of expansion in either cylinder is its total ratio of 
expansion divided by the product of all the actual expan- 
sions in the cylinders preceding it. 

The terminal pressure in either cylinder is found by 
dividing the initial pressure by the total ratio of expan- 
sion for that cylinder. 

To FIND THE Part of Stroke in Expansion, divide 
1 by the total or actual ratio of expansion, as is required, 
and subtract the quotient from 1. The advantage of 
a compound engine is in allowing a much greater ratio 
of expansion with a more equal distribution of the heat 
in the steam than can be obtained in a single engine. 



THE engineer's EPITOME. 77 

The water consumption is calculated from the termi- 
nal pressure in the last cylinder, as is also the amount of 
water for condensation. 

The average pressure in either cylinder is found from 
the part of stroke in expansion found by the total ratio 
of expansion for that cylinder. 

To FIND THE Minimum Pounds of Water to con- 
dense 1 Pound of Steam, Jet Condenser. — Rule: 
From the total heat of the steam subtract the tempera- 
ture of the water in the condenser minus the temperature 
of condensing water. 

The amount of cooling surface in a surface condenser 
should be from § to | the amount of heating surface in 
the boilers to produce the steam condensed. The effi- 
ciency will depend upon its form and the velocity at 
which the cooling water is passed through it. The water 
and steam should pass through the condenser in oppo- 
site directions. 

The diameters of cylinders in compound engines and 
cut-off should be so proportioned that each cylinder shall 
develop about an equal power. 

To FIND the H. p. of a Compound Engine from 
the Initial Pressure of the High Pressure Cyl- 
inder Card and the Terminal Pressure of the 
Low Pressure Cylinder Card. — Rule : Divide the 
initial pressure of the high pressure card by the termi- 
nal pressure of the low pressure card. The quotient is 
the total ratio of expansion, from which and the initial 
pressure the total H. P. can be calculated from the low 
pressure cylinder as before. 

Example. — The diameter of the low pressure cylinder 
is 70 inches ; stroke, 66 inches ; revolutions, 90. The 
initial pressure of high pressure cylinder card is 120.3 



78 THE engineer's epitome. 

gauge ; terminal pressure low pressure cylinder card, 4.5 ; 
absolute back pressure in condenser, 2.5. What is the 
H. P. of the engine ? 

120.3 + 14.7 = 135 

135 -r 4.5 = 30 = total ratio of expansion. 

1 ^ 30 = .033 

1 —.033 = .967 

.967^= .9351 

.033 X 1.8 = .059 

1.03 -\- .059 = 1.089 

.9351 -i- 1.089 = .858 

1 —.858 = .142 

135 X .142 = 19.17 = average pressure. 

19.17 — 2.5 = 16.67 = effective pressure. 

70^ = 4900 

4900 X 66 = 323400 

90 X 323400 = 29106000 

4 X 29106000 = 116424000 

16.67 X 116.424 = 1940.78 

14.44 = correction. 

1940.78 — 14.44 = 1926.34 = total H. P. 

The point of cut-off can be found from the average ■ 
pressure by the following rule : Divide the average press- 
ure by the initial minus the average pressure. Add 1.74 
to the quotient, and extract the square root of the sum. 
Divide 1.34 by this root. Subtract the quotient from 1. 
Multiply the remainder by the stroke. 

Formula : s / 1 — ■ . \ 

V A'^1- + 1.74/ 
p — a 

Example. — The initial pressure is 100 pounds. The 

average is 40 pounds ; stroke, 50 inches. What are the 

point of cut-off, ratio of expansion, and terminal pressure ? 

100 — 40 = 60 
40 ^ 60 = .666 
.6 66 + 1.74 = 2.406 
-J2.4O6 = 1.551 
1.34 -!- 1.551 = .863 



THE engineer's EPITOME. 79 

1_. 863 = .137 

50 X .137 = 6.85 = cut-off. 

1 — .136 = 7.34 = ratio of expansion. 

100 -r 7.34 = 13.62 = terminal pressure. 

The best results from a compound engine are obtained 
when the cut-off valves to the intermediate and low press- 
ure cylinders are so proportioned as to give a pressure in 
each receiver equal to the terminal pressure in the pre- 
ceding cylinder, as, in order to maintain a higher press- 
ure than the terminal in the receiver, the cut-off in the 
succeeding cylinder must be less, thereby reducing the 
average and increasing the back pressures. There will 
be a loss of power in the high pressure and all the inter- 
mediate cylinders by increasing the pressure in the re- 
ceiver beyond the terminal pressure of the preceding 
cylinder, while there will be an increase of power in 
the low pressure cylinder. On the contrary, by making 
the receiver pressure less than the terminal, there is an 
increase of power in the high pressure and intermediate 
cylinders, with a decrease of power in the low pressure 
cylinder. This decrease of pressure in the receivers can 
be carried to such an extent as to make the low pressure 
cylinder useless. The rule for finding the ratio of ex- 
pansion in either cylinder, when the pressure in the pre- 
ceding receiver is given, is as follows : Divide the square 
of the diameter of the succeeding cylinder multiplied by 
the pressure in the preceding receiver by the square of 
the diameter of the high pressure cylinder multiplied by 
its cut-off in decimals of stroke and by the initial pressure. 

To show the effect of receiver pressures, suppose the 
diameters of the cylinders to be 20", 40", and 80" ; ratio 
of expansion in high pressure cylinder, 3 ; absolute ini- 
tial pressure, 160 ; revolutions, 100 ; stroke, 60" ; absolute 
back pressure, 2. First, suppose 5 pounds' pressure in 



80 THE ENGrlXEER's EPITOME. 

each receiver above the terminal pressure due to expan- 
sion ; terminal pressure in high pressure cylinder, 5o^; 
average, 111.3 ; effective pressure, 111.3 — 53.3 — 5 = 53; 
H. P. high pressure cylinder, 508.8 ; pressure in first re- 
ceiver, 58.3; ratio expansion intermediate cylinder, 4,37; 
terminal pressure, 13.3 ; average pressure, 32.55; effective 
pressure, 32.55 — 13.3 — 5 = 14.25 ; H. P. intermediate cyl- 
inder, 547.2 ; pressure in second receiver, 18.3 ; ratio ex- 
pansion low pressure cylinder, 5.5; average pressure, 
8.99; effective pressure, 8.99 — 2 z= 6.99 ; H. P. low press- 
ure cylinder, 1073.8 ; total H. P., 2129.8. 

Again, suppose 5 pounds less in each receiver than 
that due to terminal pressure. Terminal pressure in 
high pressure cylinder, 53^; average, 111.3; effective, 
111.3 — 53.3 + 5 = 68; H. P. high pressure cylinder, ^ 
604.8 ; pressure in first receiver, 48.3 ; ratio expansion , 
intermediate cylinder, 3.62 ; average pressure, 30.4 ; ter- 
minal pressure, 13.35; effective pressure, 30.4 — 13.35 + 
5 = 22.05; H. P. intermediate cylinder, 852.7; pressure 
in second receiver, 8.35 ; ratio of expansion low pressure 
cylinder, 2.51; average pressure, 4.73; back pressure, 2; 
effective pressure, 2.73 ; H. P. low pressure cylinder, 
419.3 ; total H. P., 1876.5. Total H. P. calculated from 
each cylinder, pressure in each receiver equal to terminal 
pressure in preceding cylinder, 2160.3. Calculated from 
low pressure cylinder only, 2199.5 H. P. Having been 
worked very accurately, the last two should agree. The 
last, however, is the one most likely to be correct. 

The effective pressure on either cylinder of a compound 
engine, except the last, is very nearly the same for any 
part of cut-off between .2 and .6 of the stroke, being 
about .34 of the initial pressure in that cylinder. In the 
low pressure cylinder the effective pressure between the 
above cut-offs will be about .34 of the initial pressure for 



THE engineer's EPITOME. 81 

that cylinder plus the difference between the terminal 
and the back pressures. 

Find the relative diameters of the cylinders in com- 
pound engines, so that there shall be an equal power 
exerted by each cylinder, the pressure in the receivers 
being equal to the preceding terminal pressures, by the 
following rule : Raise the ratio of expansion in the high 
pressure cylinders to a power one less than the number of 
the cylinder, and extract the square root. This root will 
be the relative diameter, the diameter of the high press- 
ure cylinder being 1 ; or, in three cylinder engines, the 
diameter of the third cylinder equals the ratio of expan- 
sion in the high pressure cylinders. The diameter of the 
intermediate cylinders equals the square root of the ratio 
of expansion. 

Example. — The ratio of expansion in the high pressure 
cylinder being 3 and its diameter 10, what should be the 
diameters of the succeeding cylinders, so that each may 
develop an equal power? The diameters are as the 
square roots of 1, 3, 9, 27, 81, 243, etc., or 10, 17.2, 30, 
51.9, 90, 155.9, etc. The effect of clearance in the cyl- 
inders will probably vary the equality of the above 
results a little. 



82 THE engineer's epitome. 



THRUST ON PROPELLER. 



Rule for finding the number of feet passed over by a 
vessel per minute : Multiply the knots per hour by 101.3. 

To FIND THE Calculated Speed per Hour in 
Knots, use the following rule: Multiply the pitch of 
propeller in feet by revolution per minute and by .00987. 

To FIND THE Thrust of Propeller in Pounds, use 

the following rule: Multiply the diameter of propeller 
wheel by 4..5. From the product subtract the pitch of 
wheel multiplied by 1.1. Multiply the remainder by the 
diameter of the wheel and by the calculated knots per 
hour plus the slip. The slip is the difference between 
the calculated knots and the knots actually run. 

Formula for thrust: D (4.5 D — 1.1 P)x (.00987 P R 
+ S). D = diameter of wheel, P=r pitch, R = revolution 
per minute, and S = slip. 

To find the H. p. of Thrust. — Multiply the pounds 
at thrust, as found above, by .00003 times the revolution 
per minute multiplied by the pitch. This H. P. is usu- 
ally but about .40 per cent, of the actual power of the 
engine. 

Example. — The diameter of wheel is 24 feet, pitch 30 
feet, revolutions 70, actual run 19.25 knots. What is the 
calculated speed, thrust, and H. P. ? 

30 X 70 X .00987 = 20.73 knots. 

20.73—19.25 = 1.48 = slip. 

20.73 + 1.48 = 22.21 

24 X 4.5 — 30 X 1.1 =- 75 X 24 = 1800 X 22.21 = 39977 = 

thrust. 
30 X 70 X .00003 = .063 X 39978 =- 2518.6 = H. P. at thrust. 
Or 2518.6 -I- .40 = 6296.5 = H. P., actual. 



I 



THE ENGINEER S EPITOME. 



SALT WATER FEED.— Jet Condenser. 



The amount of salt in sea water varies considerably in 
different places, but is generally considered as -^^ of its 
weight in salt. At about ^f of its weight in salt, water 
becomes saturated and will take up no more. The 
saltest natural bodies of water known are the Dead Sea 
and Great Salt Lake, each of which contains about 3^3 of 
its weight in salt, which is one-half the saturation point. 
The temperature at which water boils varies somewhat, 
according to its degree of saltness. In using salt water 
feed for boilers, the quantity blown from the boiler must 
always be such a part of the quantity pumped in as is 
denoted by the number of times the thirty-thirds of salt 
in the boiler is contained in the thirty-thirds of salt in 
the feed water. Thus suppose the sea water contains 
li thirty-thirds of salt, the limit at which the boiler is to 
be salted being 2^ thirty-thirds. Then 1^ divided by 2^, 
which equals -|, is the part of feed that must be blown 
from the boiler to maintain that limit of saltness, and 
but I of the water pumped is evaporated. 

Rule for finding the percentage of loss from blowing 
off sea water feed : Divide the temperature of the steam, 
minus the feed, multiplied by the thirty- thirds of salt in 
the boiler, minus the thirty-thirds of salt in the feed 
water, plus the temperature of the steam, minus the tem- 
perature of the feed. 

The percentage of increase of feed is found by divid- 
ing 1 by the difference in the saltness in the boiler and 
the feed water. 

Example. — Saltness in the boiler is two and a half times 



84 THE engineer's epitome. 

that in the sea water ; the gauge pressure is 90 pounds ; 
temperature of feed, 135°. What is the per cent, of 
loss by blowing water from the boiler to maintain that 
degree of saltness, and what is the per cent, of increase 
of feed? Ansicer: Per cent, of loss .0976; increase feed 
.66f per cent. 

The per cent, of feed blown will be decreased a little 
by taking the feed from the hot well, as the amount of 
saltness will be decreased in the proportion of weight of 
injection water to weight of steam condensed. Thus 
suppose 20 pounds of water are required to condense 
one pound of steam : the proportion of salt in the feed to 
salt in the sea water will be reduced. 

Example. — The sea water is ^^ salt. The limit of salt 
in the boiler is g^g. The weight of injection water to 
steam condensed is 14.5 to 1 ; temperature condenser, 
140° ; sea water, 70. What part of the water pumped to 
the boiler is blown off ? What is the per cent, increase 
of feed ? What is the total heat in the condenser and 
the vacuum ? 

Rule for finding the total heat in the condenser : Mul- 
tiply the difference of heat between the condenser and 
feed water by the number of pounds of water required 
to condense one pound of steam. To the product add 
the temperature in the condenser. 

Formula : N (T — f ) + T, in which N indicates num- 
ber of pounds of water to condense one pound of steam ; 
T, temperature in condenser ; f , temperature feed. 

To FIND the Absolute Pressure in the Conden- 
ser ABOVE Two Pounds, use the following rule: 
Subtract 1150 from the total heat. Divide the remainder 
by 8. To the square of the quotient add 1.9. 

/T — 1150\2 
Formula: ( ^ ) +1.9, in which T is total 

heat in condenser or in vacuum. 



THE engineer's EPITOME. 85 

Proportion of injection water to steam, 14.5 ; propor- 

tion 01 salt m feed to sea water, ; x 4 = — = 

15.5 15.5 93 

14 5 
part of feed blown off ; 3 — —^ = 2.06 ; 1 -^ 2.06 = .485 

15.5 
= per cent, increase feed; (140 — 70) x 14.5) + 140 = 

11 55 1150 

1155 = total heat; and = .62.5. .Q25'--\-1.9 = 

8 
2.29 = absolute pressure. 14.7 — 2.29 = 12.41 = vacuum. 



THE ENGINEER S EPITOME. 



SLIDE VALVES. 



A slide valve is generally so proportioned as to give a 
definite point of cut-off of steam to the cylinder and of 
release of steam from the cylinder. In order that steam 
shall be cut off before the end of the stroke, the length 
of the valve must be greater than the distance between 
the outer edges of the steam ports. This additional 
length of valve is called "outside" or "steam lap." 
That the steam may be held in the cylinder longer and 
compressed more, the inner or exhaust edges of the 
valve are frequently made at a less distance apart than 
the inner edges of the steam ports. This is called 
"inside" or "exhaust lap." Thus aa and da (Fig. 1) 
are outside lap : hh and hh are inside lap. The stroke 



Fig. 1. 

of valve to give full open ports must equal twice the 
steam lap aa plus the width of the port ah. To admit 
steam at the commencement of stroke of engine, with 
varying stroke of valve, it is necessary that the eccentric 
should move backward and forward and across the shaft. 
This is done in some of the automatic slide-valve engines. 
In Fig. 2 are shown the positions of the eccentric to 
open the ports at commencement of stroke of engine, 
with varying stroke of valve, c is the centre of the 



THE ENGINEER S EPITOME. 



87 



shaft ; a^, a^, a% a\ and a^ are positions of centre of 
eccentric for different lengths of strokes of valve ; c^, c^, 
c^, c^ and c^ equal one-half the lengths of stroke; h^, h^, 
h'^, h\ and ¥ are positions of extreme throw of eccentric. 
The line ccv" equals one-half the line cr^ which equals 
one-half of the outside lap plus the lead. The lines co}, 
ca-, ca^, ca'^, and ca^ indicate the distance from the 
circumference of the eccentric circle to the points 6\ b^, 




Fig. 2. 

h^, ¥, and b^, equal to the distance oi the centre of the 
eccentric from the shaft, which equals one-quarter the 
stroke of the valve. Here it is seen that the extreme 
throw of the eccentric follows a curved line, while the 
centre of the eccentric follows a straight line perpen- 
dicular to the centre of the shaft, and at a distance from 
it equal to one-half the lap plus the lead. A slide valve 
working in the manner here described should have an 
independent exhaust valve. 



bo THE ENGINEER S EPITOME. 

To FIND THE Angular Advance of the Eccen- 
tric FOR ANY Lap and Stroke of Valve. — Rule: 
From the square of one-half the stroke of valve sabtract 
the square of the lap plus the lead. Divide the square root 
of the remainder by one-half the stroke. The quotient 
equals the cosine of the angular advance. Subtract the 
cosine from 1, divide the remainder by .0001535 minus 
.00003 of the remainder, and extract the square root of 
the quotient. The root equals the angular advance. 
The angular advance of the eccentric is the angle 
between a perpendicular to the centre of the shaft and 
the eccentric. 

Formula : 

V -OS 

gular advance, in which s indicates stroke of valve and 
1 the lap plus the lead. 

Examples. — The stroke of the valve is 6 inches ; the 
lap, 1| inches; lead, J^ inch. What is the angular 
advance of eccentric? 

H + sV = HI' or 1.53 inches = lap and lead. 
1.53-^ = 2.34 = square of lap plus lead. 
9 — 2.34 = 6.66 



" ^/ V.0001535 — .00003 (1 — c) / 



V^ 



J6.66 = 2.58 
2..58 -^ 3 ^= .86 = cosine. 
1 —.86 =- .14 
.14 — .00003 = .0000042 
.000153.5 — .0000042 = .0001493 
.14 -^ .0001493 = 937.7 
-1/937.7 = 30.6 = angular advance. 

The effects of inside lap are later release and earlier 
compression. The angle formed at the centre and be- 
tween extremities of points of inside and outside laps 
equals the angle at which release takes place before the 
end of stroke. 180° minus the cut-off in degrees, di- 



THE ENGINEER S EPITOME. 89 

vided by 2, equals the angular advance : hence cut-off 
equals 180° minus twice the angular advance. Com- 
pression begins at twice the angular advance of eccentric 
minus the release, before the end of the stroke. 

To FIND THE Lap of Yalve to Cut-off at any 
Angle of Revolution. — Rule : Subtract the required 
cut-off from 180°. Multiply the remainder by 86800. 
Subtract from the product the sum of the cube and the 
square of the remainder. Multiply the last remainder 
by one-half the stroke of the valve. Cut off seven 
figures. 

(N 86800) — (N3 + W) 

Formula: A s ^— ^ ^^ 

10000000 

in which N indicates 180° minus cut-off, and s stroke of 

valve. 

Example. — The stroke of valve is 8 inches, and it is 
required to cut the steam from the cylinder at ^ of the 
revolution. What lap must be given the valve, and what 
is the angular advance of the eccentric ? 

I of the revolution equals 60°. 

180 — 60 = 120 

120 X 86800 = 10416000 

120' =14400 

1203= 1728000 

14400 + 1728000 = 1742400 

10416000— 1742400 = 8673600 

8673600 X 4 = 34694400 

34694400 ^ 10000000 = 3.46944 = lap. Answer. 

To find Inside Lap to Release at any Angle 
before the End of Stroke, — Rule: From 90° sub- 
tract one-half the angle at cut-off and the angle at which 
release is wanted. Then proceed as in the rule and 
formula above. 

Example. — The stroke of valve is 5 inches. The steam 
lap plus the lead is 2 inches. The angle at which release 



90 THE engineer's EPITOME. 

is to take place before the eud of stroke is 12°, What 
are the angular advance of eccentric, points of cut-off? 
and coEQpression in degrees and inside lap ? 

5 -f 2 = 2.5 

2,5^ = 6.25 

6.25 — 4 = 2.25 

a/2T5 = 1.5 

1.5-^2.5 = .6 

1— .6 = .4 

.0000.3 X .4 = .000012 

.0001 535 — .00001 2 = .000141 5 

.4 -^ .0001415 = 2826 

'i,''2826 = 53.1 = angular advance. 

2 X 53.1 = 106.2 

180 — 106.2 = 73.8 = cut-off. 

36.8 = 1 cut-off. 

90—36.8 = 53.2 

12 = release. 

53.2— 12 = 41.2 = N. 

41.22=1697.44 

41.2 X 1697.44 = 69934.52 

1697 + 699.34.52 = 71631.52 

41.2X86800=3576160 

3576160—71631.52 = 3504529 

3504529 -f 10000000 = .3504 

21- X .3504 = .8760 = mside lap. 

106.2 — 12 = 94.2 = compression from end of stroke. 

180 — 94.2 = 85.8 compression from beginning of stroke. 

36.8 = I cut-off. 

90 — 36.8 = 53.2 

12 = release. 

53.2 — 12 = 41.2 =N. 

41.2 X 86800 = 3576160 

41.2^ = 69934 

41.2-2 = 1697 

69934+1697 = 71631 

3576160—71631 = 3504529 

3504529 X 2^ = 8761322 

8761322 -^ 10000000 = .8761322 = inside lap. 

To FIND Lap to Cut-off at any Part of Stroke 
OF Engine. — Rule : From 1 subtract the cut-off divided 
by the stroke. Divide the remainder by 4 minus one-half 



THE engineer's EPITOME. 91 

the cut-off divided by the stroke. Multiply the square root 
of the quotient by stroke of valve. To the product add 
one-half the lead. This rule gives the lap where the 
connecting rod is four times the length of the crank. 

Example. — The length of connecting rod is 65 inches ; 
crank, 10 inches ; cut-off at 6 inches of stroke ; lead, ^ 
inch ; valve stroke, 7 inches. What is the lap ? 

[1 — (6 -^ 20)] ^ [4 — (3 -^ 20)] X 7 + * lead = lap. 

6 ^ 20 = 3 

1 — .3 = 7 

3 -^20 = .15 

4 — .15 = 3.85 

7^3.85 = .1818 

-J. 1818 = .424 

.424 X 7 = 2.968 

ilead = ^2 = .03125 

2.968 + .03125 = 2.999 lap required. 

Lap given, to find Cut-off. — Rule : From the square 

of the stroke of the valve subtract four times the square 

of the lap minus the lead. Divide the remainder by the 

square of the stroke of the valve minus one-half the square 

of the lap minus the lead, and multiply the quotient by 

the stroke of the engine. 

o s2 — 41 
Formula : S 5 

s' — if 

in which S indicates stroke of engine ; s, stroke of valve ; 
and 1, lap of valve minus the lead. 

Example. — The stroke of the engine is 42 inches ; 
valve stroke, 6 inches; lap, 1.47; lead, Jg inch. What 
is the distance from commencement of stroke at which 
cut-off takes place? 

Lap plus I lead equals 1.5, then 



4 X 2.25 = 9 
36 — 9 = 27 



U2 THE ENGINEER S EPITOME. 

2.25-^2 = 1.125 

36—1.125 = 34.875 

27-^34.875 -=.774 

42 X .774 = 32.508 = cut-off. Answer. 

To FIND THE Width of Port Opening at any 

Angle of Cut-off. — Rule : From 800 times the square 

of the angle at cut-off subtract the cube of angle at 

cut-off. Multiply the remainder by one-quarter the 

stroke of the valve, and point off seven figures. 

, .25 s (800N2 — N3) 

Formula: 

10000000 

in which N indicates the angle at cut-off, | = .25. 

Exajnple. — The cut-off is at 100°. The stroke of the 

valve is 4 inches. How much is the port opening? 

1002=10000 

800 X 10000 -= 8000000 

1003 = 1000000 

8000000— 1000000 = 7000000 

i of 4 = 1 

1 X 7000000 = 7000000 

7000000 -^ 10000000 = .7 inch. Answer. 

To FIND THE Width of Port Opening when the 
Cut-off is at any Part of Stroke and when the 
Cut off is Less than One-half Stroke — Rule : 
Multiply twice the cut-off plus the square of the cut-off 
by .11 of the stroke of valve. 

Formula: .lis (2c + c^) 

Where the cut-off is greater than one-half the stroke, 
use the following rule: Multiply the square root of 1 
minus the cut-off by 1.025. Subtract the product from 
1, and multiply the remainder by one-half the stroke of 
the valve. 

Formula : ^ s (l — 1.025 V 1 — c) 

Examples. — The stroke of engine is 36 inches ; cut-off 
at 15 inches ; valve stroke, 6 inches. What is the width 
of port opening ? 



THE engineer's EPITOME. 93 

15-^36 = .416 

.41 6-^ = .173 

.173 + .834 =1.007 

.66 X 1.007 = .664 inch. Ansiver. 

The stroke of engine is 42 inches ; cut-off at 28 inches ; 
valve stroke, 7 inches. What is the width of port open- 
ing ? Generally, the width of port opening equals one- 
half the stroke of the valve minus the lap. 

28 ^ 42 = .6666 

a/.6666 = .816 

1.025 X .816 = .8364 

1 — .8364 = .1636 

3.5 X .1636 = .5726 inch. Ansiver. 

The area of a steam port equals the square of the 
diameter of the cylinder multiplied by the distance the 
piston travels, in feet, per minute and by .000131. 

To set a slide valve, adjust the length of the valve 
connecting rod, so that the valve uncovers the ports an 
equal distance at each end. Then, if the builders of the 
engine have keyed the eccentric to the shaft, the valve 
is supposed to be " set." If the eccentric is not keyed to 
the shaft, place the engine on the "centre," turn the 
eccentric ahead the way the engine is to run till the 
port at the end on which the piston is, just begins to 
open. Turn it a little farther ahead until you get the 
" lead " you require. If the compression is strong, but 
little " lead " is required. Turn the engine on the oppo- 
site centre, and notice if the opposite port opening is the 
same as the other. If not, continue the adjustment until 
it is. An engine does not always work best when the 
valves are equally adjusted. A little more opening is 
generally given to the lower port than the upper in 
upright engines. This, however, depends on the counter- 
balance of the crank and moving parts. Link motion 



94 THE engineer's epitome. 

reversing gear is usually used on locomotive and marine 
engines. 

To set the vaive on a link motion engine, place the 
reverse lever in the position in which it is to be ordi- 
narily used, both for ahead and back, and adjust for 
either place, as directed above. Two or three trials may 
be needed on account of distortion, caused by neither 
being at first in its proper place. Many slide valve 
engines, especially marine engines, have an adjustable 
cut-off valve working on the back of the main valve, the 
adjustment being made at any time by means of a right 
and left screw, bringing the cut-off valves to the position 
desired. The eccentric for this valve is placed opposite 
the crank. 



THE ENGINEER S EPITOME. 



95 



GRAPHIC METHOD FOR FINDING 
THE W^ORKING AND PROPOR- 
TIONING OF A SLIDE 
VALVE. 



On the line ss (see fig.)> equal to the stroke of the 
valve, draw a circle. From the centre c set off ci, 
equal to inside lap, co, equal to outside lap, and ob, 
equal to lead. Through the points ioh draw lines per- 




pendicular to ss, touching the circle. Through the 
point where b touches the circle draw ce. e is the 
position of extreme throw of the eccentric. Open a 
pair of compasses from e to n, then, with one point on 
s, extend the other to d: d is the position of crank at 
cut-off. Open compasses from s to e, place one point 
on r, extend the other backward to r. This is the posi- 
tion of crank at release. Place one point on w, extend 
backward to n. n is the position of crank when compres- 



9b THE ENGINEER S EPITOME, 

sion begins. Make aa equal to stroke of engine on any 
convenient scale, draw the circle to represent the path of 
crank pin, continue all the lines from the centre to this 
circle, and extend the line aa indefinitely beyond a. 
Open the compasses to represent connecting rod at the 
same scale, place one point on release, the other on 
extended line, and draw the curve from release to 1. 
With one point on cut-off and the other on extended 
line, draw cut-off 2, and same for compression 3. 1 is 
the position of crosshead when exhaust opens before the 
end of stroke, 2 is the position of crosshead when steam 
is cut off, and S is position of crosshead when exhaust 
closes or compression begins. The difference between 
the lap and half stroke of valve is the width of port 
opening. By altering the stroke and making another 
stroke circle, the lap and lead remaining the same, or by 
altering the lap and lead, the stroke remaining the 
same, the different positions of cut-off, release, compres- 
sion, and eccentric can be shown. 

To reverse the engine, place the eccentric on the lower 
outer quarter. If the valve works with a reverse motion 
arm or takes steam inside, place the eccentric on the 
inside lower quarter. The inside and outside lap can be 
found from the figure, when the desired cut-off and 
release are given. 



THE ENGINEER S EPITOME. 



97 



RELATIVE POSITIONS OF PISTON 
AND CRANK. 



It may at times be convenient to determine the part of 
revolution the crank has made when the piston or cross- 
head has made a definite part of the stroke, or, having 
a certain part of the revolution given, to determine the 
distance the crosshead has moved. 

To FIND THE Angle of Revolution from the Dis- 
tance THE Crosshead has moved. — Rule : Divide 
the length of connecting rod by the length of crank. 




Call the quotient r. Divide the distance crosshead has 
travelled by length of stroke. Call the quotient N. 
Then, for first half of forward stroke, or N minus 1, and 
180 minus angle of return stroke, the formula is : — 



\/^C 



12650- 



7360\ 



(36 + ^r) m = 



ang. 



le. 



For first half of return stroke, or 1 minus jSJ", and 180 
minus angle of forward stroke : — 
Formula : — 



y/ X (l2650 + ?^') + (7 + 2 r) N"^ = angle. 



90 THE ENGINEEK S EPITOME. 

Angle given, to find part of stroke, first quarter revo- 
lution forward stroke, or 180 minus angle, and 1 minus 
N, return stroke : — 

a^ c- 
Formula : = c : c = X. 

12850-1552 « 

r 

For first quarter revolution return stroke, or 180 minus 

angle, and 1 minus N, forward stroke : — 

Formula : = c : c — c2 (9 +1^ r) = N. 

24400 

12650 ^ i- 

^ r + 1 



THE ENGINEER S EPITOME. 



STRENGTH OF THE PRINCIPAL 

WORKING PARTS OF AN 

ENGINE. 



To FIND THE Thickness of the Castings for 
Cylinder. — Rule: Multiply the diameter of the cylin- 
der by the absolute pressure. To the product add 1500 
times the square root of the diameter of cylinder, and 
point off four figures from the right side. 

Formula: (DP) + 1500^17 
10000 
in which P indicates absolute pressure and D diameter 
of cylinder. 

Example. — The diameter of the cylinder is 40 inches. 
The gauge pressure is 65 pounds. What should be the 
thickness of castings of cylinders ? 

65 + 15 = 80 = absolute pressure. 
40 X 80 = 3200 
-^40 = 6.32 
1500X6.32 = 9480 
9480 + 3200= 12680 

Cutting off four figures, or, in other words, dividing by 10000, 
makes the result 1.268 = thickness of casting. 

To FIND THE Length of Crank Pin. — Rule : Mul- 
tiply the diameter of the cylinder by the square root of 
the absolute pressure and by .0365. 
. Formula: .0365(D-^"P) 

Example. — Diameter of cylinder is 90 inches ; abso- 



100 THE engineer's EPITOME. 

lute pressure in cylinder, 16 pounds. What should be 
the length of crank-pin? 

4 X 90 = 360 

360 X .0365 = 13.14 inches. Answer. 

To FIND THE Diameter of Crank Pin, — Eule : 
Multiply the diameter of the cylinder by the square root 
of the absolute pressure and by .0265. 

Formula: .0265 (D^/P") 

Example. — What should be the diameter of crank pin 
in the last example ? 

360 X .0265 = 9.54 inches. Ansiver. 

In compound engines the diameter of pin for second 
engine should be increased by 1.2 of the above rule, and 
the third engine pin by 1.4. 

To find the Diameter of Piston Rod. — Rule: 
Multiply .015 plus .00004 S by the diameter of cylinder 
multiplied by the square root of the absolute pressure. 

Formula: D^'T (.015+ .00004 S) 
in which S indicates stroke of engine. 

Example. — The stroke in the above example is 60 
inches. What should be the diameter of piston rod ? 

60 X .00004 = .0024 
.015 4- .0024 = .0174 
360 X .0174 = 6.26 inches. Ansiver. 

To find the Diameter of Connecting Rod. — 
Rule : Multiply the stroke by .00005. Then proceed as 
in finding piston rod. For steel rods, multiply the above 
results by .8. 

To find the Diameter of Shaft. — Rule: Multiply 
the square of the diameter of cylinder in inches by stroke 
in inches, by absolute pressure, and by .00024, and extract 
the cube root of the product. 



THE engineer's EPITOME. 101 

Formula: ^'d^ x S x P x .00024 

In compound engines, multiply the above result by 1.2 
for second cylinder, and by 1.4 for third cylinder. This 
rule gives the minimum diameter of the shaft. For steel 
shafts, multiply the diameter for wrought iron by .8. 
The total weight or pressure on the crank pin divided 
by .00014 of the weight should not exceed 8000 pounds. 
The pressure on a pin or bearing should not exceed 800 
pounds per sectional inch. 

The above rule for shafts gives about I the total tor- 
sional strength of the shaft. 



102 



THE ENGINEER S EPITOME. 



WATER. 



The motion or flow of water is caused by gravity, and 
follows the same laws as falling bodies. The theoretical 
velocity of water discharged from an orifice in the bottom 
or side of a vessel is the same as the velocity of a body fall- 
ing by gravity from a height 
77^ r\ ^ ^.00 of the water above the cen- 

tre of the orifice. The theo- 
retical velocity is 8.02-5 times 
the square root of the height 
above, or 12.5 times the press- 

f^- H- Zd'O ^^^ P^^ square inch at the 

S" ^ orifice. This velocity is sel- 



0— > '■'' 



^.<J7 



^^' dlD- 



^■V^ 



^.1^0 



dom or never obtained in 
practice, owing to the interfer- 
ence of the particles of water 
in rushing into the orifice, 
with the proper multiples for 
each. The velocity found from 
the pressure is subject to the 
same reduction for form of 
orifice as that found from the 
height. 
To FIND THE Cubic Feet of Water discharged 

THROUGH AN OrIFICE IN THE BOTTOM OR SiDE OF A 

Vessel. — Kule : Multiply the square root of height 
above the centre of the orifice, when the orifice is in 
the side, or the square root of the height above the 
bottom, when the orifice is in the bottom, by .327 times 



Fig. 1. 

Form of orifices and multi- 
pliers lor pressures with dif- 
ferent heads of water. These 
multipliers can also be used 
for square orifices of same 
general outlines. 



THE p:ngineer's epitome. 103 

the square of the diameter in inches when the orifice is 
round, or by .416 times the area of the orifice and by 
multiplier corresponding to form of orifice. 

Formulas: .327D2m'Yp when orifice is round, or 
,416 A m a/ P when orifice is rectangular. 

When the pi'essure at the orifice is given, the formulas 
are as follows : — 

.327D2ra^p~ and .416 A m>J p~ 
in which m indicates multiplier corresponding to form 
and pressure. 

Example. — The head of the water above the centre of 
orifice is 50 feet; the orifice is circular, 3 inches in diam- 
eter, in a thin sheet. What is the velocity of discharge 
in feet per second, and what quantity of water could be 
discharged per minute ? The number corresponding to 
form of orifice is 5. 

-^50^7.07 

5 X 7.07 = 35.35 = velocity. 

32 = 9 

:327 X 9 = 2.943 

7.07 X 2.943 = 20.807 

5 X 20.807 = 104.03 cubic feet per minute. 

The floor of a wooden tank is 4 inches thick; the 
opening in the bottom is 2 inches square ; pressure on 
bottom, 20 pounds per square inch. What is the velocity 
of discharge in feet per second, and how many cubic feet 
are discharged in one minute ? ° The number correspond- 
ing to form of orifice and pressure is 10. 

-y}2Q = 4.47 

4.47 X 10 = 44.7 = velocity 

.416 X4=: 1.664 

1.664X10 = 16.64 

16.64 X 4.47 = 74.38 cubic feet of water discharged per minute. 



104 THE engineer's EPITOME. 

To FIND THE Time in ^yHICH a Vessel will 

EMPTY itself THROUGH AN OrIFICE IN THE BoTTOM, 

OR TO THE Bottom of an Orifice in its Side. The 
theoretical average flow with which a vessel will empty 
itself is 5.39 times the height of the fluid above the ori- 
fice multiplied by the square root of the height plus 1.25, 
divided by the height plus 1. Then, to find the time, 
use the following rule : Multiply the cubical contents of 
the vessel in feet by the height plus 1. Divide the prod- 
uct by .28 times the area of orifice multiplied by height, 
multiplied by square root of height plus 1.25, multiplied 
by number corresponding to form of orifice. 

c(h + l) 



2^ X A X h X ^h + 1.25 
c(h+l) 



The theoretic formula is : 

and practical formula : 

.28 X m X A X h X ^h + 1.25 

in which c indicates cubical contents in vessel, A area of 
orifice, and m number corresponding to form of orifice 
and head. 

Examples. — The height of water in vessel is 30 feet. 
What is the average theoretical velocity until the vessel 
is empty ? 

30 +1. 2.5 = 31.25 

-^31,25 = ,5.59 

5.59X5.39 = 30.13 

30.13 X 30 = 903.9 

903.9 ^ 31 = 29.15 feet per second. 

The flooring of a tank is 5 inches thick ; the head of 
water is 20 feet; cubic feet in tank, COCO; orifice in 
bottom of tank, 3 inches square. In what time will the 
water be discharged from the tank ? 

The multiplier corresponding to orifice is nearest 6.6. 
Then 



THE engineer's EPITOME. 105 

6000X21 = 126000 

20+ 1.25 =21.25 

AI21.25"= 4.6 

4.6 X 20 -= 92 

92 X 9 = 828 

828 X .28 = 231.84 

231.84X6.6 = 1.5.30.144 

126000 -^ 1530.144 = 82 minutes. 

To FIND THE Height of Water above the 
Centre of Orifice to discharge a Given Quan- 
tity OF Water per Minute. — Rule : Divide the 
quantity discharged by .416 times the area of orifice 
multiplied by number corresponding to form of orifice. 
Square the quotient. Or, for a second orifice, divide the 
quantity discharged by .327 times the diameter of orifice 
multiplied by number corresponding to form of orifice. 
Square the quotient. 

Formula : ( — — ) , or, for circular orifice, 

\.416 X A X m/ 



\.327 X D2 X m/ 

Example. — The number of cubic feet discharged per 
minute is 400 ; the diameter of discharge pipe is 4 inches; 
length, 8 inches. What is the head of water to give the 
discharge ? 

Multiplier, 6.6. Then 

42=16 

16 X .327 = 5.232 
6.6 X 5.232 =.34.53 
400 -I- .34.53 = 11.58 
11.582 = 1.34 = head. 

To find the Velocity and Discharge of Water 
through Long Pipes. — Rule: Divide 64.4 times the 
head, or 150 times the pressure at orifice, by 1.5 plus the 
length of the pipe in feet divided by diameter in feet 



106 THE engineer's EPITOME. 

multiplied by a factor for friction of, from, or to .05. 
Extract the square root of the quotient. The square 
root equals the velocity in feet per second. Multiply the 
velocity by square of diameter of pipe and by 47|, for 
cubic feet discharged per minute. 

Example. — The leugth of pipe is 10000 feet; diameter, 
3 feet; head, 45 feet; factor for friction, .03. What is 
the velocity of flow and number of cubic feet discharged 
per minute ? 

64.4 X 45 = 2898 

10000^3 = 3333.33 

.03 X 3333.33 = 99.9 

99.9 + 1.5 = 101.4 

2898jr 101.4 = 28.57 

■a/28.57 = 5.34 = velocity. 

32 = 9 

9 X 5.34 = 48.06 

47i X 48.06 = 2264.82f = cubic feet per minute. 

To FIND THE Weight of Water in Pipes. — Rule: 
Multiply the length of pipe in feet by the square of the 
diameter in inches and by .34. For rectangular vessels, 
multiply the length in inches by breadth in inches by 
height in feet and by .43. 

Example. — The diameter of a tank is 20 feet; the 
height, 15 feet. What weight of water will it contain ? 

20 feet = 240 inches. 

240- = 57600 

1 5 X 57600 = 864000 

.34 X 864000 = 293760 pounds. 

A simple rule for finding the approximate relative 
volume of water due to expansion by heat : Divide 250 
by 461 plus the temperature of the water. To the quo- 
tient add one thousandth part of 461 plus the tempera- 
ture of the water. 

Examp)le. — The temperature of the water is 200°. 



THE engineer's EPITOME. 107 

What is the relative volume, the volume at maximum 
density being 1 ? 

461 +200 = 661 

250 ^ 661 = .37 

1000 of 661 = .661 

.37 4- .661 = 1.03 = relative volume. 

The pressure of a liquid against any square foot of 
bottom surface equals the depth of the liquid multiplied 
by the weight of a cubic foot of the liquid. 

The pressure against a square foot of lateral surface 
at any depth below the surface equals twice the depth 
minus 1 multiplied by one-half the weight of a cubic 
foot of the liquid. The total 
pressure against the side equals 
the square of the depth multi- 
plied by one-half the weight of a 
cubic foot of the liquid. The 
average pressure is at one-half 
the depth; and the place at 
which all the pressure is concen- ^^' " 

trated, or the centre of gravity of the pressure, is at two- 
thirds the depth. 

In the triangle ABD (see Fig. 2) let AB represent the 
depth of the liquid, and BD the pressure at the bottom. 
The line AD shows the increase in pressure. The line 
at a at one-half the depth is the average pressure, and 
the line at c is the line at centre of pressure, or the line 
at which the pressure on the parallelogram EF above 
the line c will balance the pressure below c. Let h equal 
the line AB, the height or depth of the liquid ; p, the line 
BD, the pressure at the bottom. The trapezium BDch 
equals -^-^ of hp, which equals ^^g^ of the depth multiplied 
by the pressure. The point m, |- of the height from the 
surface, is the point of concentration of pressure, or centre 




108 THE engineer's E-PITOME. 

of gravity, on the point of parallelogram above the line 
C ; and n, -^ of the height from the surface, is the point 
of concentration of pressure below the line C. 

The pressure per square inch for any height of water 
is .432 multiplied by the height in feet. 

Examples. — What is the average pressure per square 
inch on the side of a vessel with 20 feet depth of water? 
The pressure on the surface is nothing. Then 

20 X .432 = 8.64 = pressure at bottom 

8.64 -f- =- 8.64 

8.64 -r 2 = 4.32 = average pressure. 

What is the pressure on a plank 1 foot wide and 5 
feet long in a flume 25 feet below the surface of water? 

1X5 = 5 square feet in plank. 

2 X 25 -- 50 

50 — 1=49 

31JX 49 = 1531.25 

5 X 1531.25 = 7656.25 pounds. Answer. 

What is the total pressure on the side of a pipe 6 
inches in diameter, height of water being 40 feet ? Also 
what is the weight of water in the pipe ? 6 inches equals 
.5 foot. Then 

3.14 X .5 = 1.57 = circumference. 

311 X 1.57 = 49.062 

] 600 X 49.062 = 78499.2 = pounds' pressure. Ansioer. 

6- = 36 

40 X 36 = 1440 

.34 X 1440 = 489.6 = weight. Answer. 

In Fig. 3, if water enters the pipe at A with a certain 
velocity, by reason of the contraction at B, the velocity 
is increased. This increase in velocity acts as a siphon, 
drawing the water from the vessel D. At C the velocity 
is less than the initial, and the water will pass out of the 
pipe into the vessel, thus, with proper proportions, form- 



THE ENGINEER S EPITOME. 



109 




Fig. 3. 



ing a circulation through the vessel. In Fig. 4 a vacuum 
is formed in the tube B by reason of the contraction of 
the stream, and the water will be drawn 
from the vessel into the tube. 

To MEASURE THE WaTER PASSING 

IN A Running Stream. — Find some 

suitable place where, for a distance of 

from 50 to 100 feet, the stream runs as 

nearly straight and of uniform width, 

depth, and velocity of current, as may 

be. Measure off a distance by the side 

of the stream as long as convenient, 

and place a stake at each end. Then 

take a piece of board loaded at one 

edge so as to be nearly submerged. 

Let it float, nearly reaching the bot- 
tom, several times through the stream 

between the stakes, placing it a few feet above the upper 

stake, and noticing the time it takes to pass between the 

stakes each time. Find the average width and depth of 
the stream by taking measurements in 
several places between the stakes. | the 
product of the average width, depth, and 
velocity per minute in feet will give the 
number of cubic feet passing the stream 
per minute. 

To FIND THE Quantity of Water 
BY Means of a River or Over- 
throw. — A dam must be made suffi- 
FiG. 4. ciently high to make " still water." The 

river over which the water flows must 
be level, and the sides perpendicular. The up-stream 
edges must be smooth and sharp, and the down-stream 




110 



THE ENGINEER S EPITOME. 



edges levered so that the water will only come in contact 
with the up-stream edges. There must be depth enough 
below the river so that the overflow will not back up and 
interfere with the velocity of discharges. The height 
or head of water is measured 
from the bottom of the river 
to the surface of the " still wa- 
ter " above the river. (Fig. 5.) 
AB = head. A is point of still 
water on surface. B = bottom 
of weir. Drive the stake AB 
and make the point B level 
with the bottom of weir. 

Rule for finding cubic feet 
discharged per minute : Multi- 
ply the length of the aperture 
in inches by the height of 
"still water" above the bot- 
tom of aperture in inches by 
the square root of the height 
and by .403. 
Formula : .403 x L x h x a/ h 

Mr. France's formula for a river of this description is 
as follows : — 

200(L — .2h)hA^/ir 
L and h are in feet, and .2 h is for contraction of stream 
at the ends of river. This formula is in general use, and 
gives about 15 feet in a thousand less than the ordinary 
rule. 

Examples. — The height of water in river is 10 inches ; 
the length of river, 8 feet. How many cubic feet are dis- 
charged per minute ? 




Fig. 5. 



THE engineer's EPITOME. Ill 

8 feet — 96 inches. 

A^10 = 3.16 

10X3.16 = 31.6 

96 X 31.6 = 3033.6 

.403 X 3033.6 = 1222.54 cubic feet. Answer. 

The height of water on river is 16 inches ; length of 
river, 10 feet. What number of cubic feet are discharged 
per minute ? 

16 inches equals 1.25 feet. Then 



^J'^ 



.25=1.118 



1.25 X 1.118 = 1.3975 

.2 X 1.25 = .25 

10 — .25 = 9.75 

9.75X1.3975 = 13.625 

200 X 13.625 = 2725 cubic feet. Answer. 

The theoretical H. P. of a stream is .0019 times the 
cubic feet discharged per minute multiplied by the head 
or height of fall. The actual power utilized depends 
very much on the useful effects the means employed for 
utilizing the power will yield. An overshot water wheel 
will give an average of about 70 per cent, of the total 
power, in useful effect, and should be used when the fall 
is 15 feet or more. The velocity of its periphery should 
be about 6 feet per second, and should have from 2 to 3 
feet head above it. A high breast wheel will yield an 
average of about 60 per cent, of the total power. It 
should be used for falls less than 15 feet, and should 
take its water a little above the centre. A low breast 
wheel takes its water below the centre, and gives an 
average yield of about 54 per cent, of the total power. 
Iron wheels, or turbines, are now in more general use 
than any other, and give better effects than the wooden 
wheels, their percentage of useful effect being all the 
way from 50 to, as claimed for some turbines, 96 per 



112 



THE ENGINEER S EPITOME. 



cent, of the total power. A probable, useful, and effect- 
ive average would be about 82 per cent. 

In testing water wheels, it may be more convenient to 
measure the water after it has passed through the wheel, 
care being taken to include only the water which has 
passed through the wheel. The measurement should be 
taken as nearly as possible simultaneously with the meas- 
urements of power from the wheel. 

Example. — The utilized head of a fall is 30 feet, the 
volume of water passing the wheel is 500 cubic feet per 
minute, the measured H. P. of the wheel is 20. What is 
the percentage of useful effect of the wheel ? 

30 X 500= 15000 

.0019 X 15000 = 28.5 = H. P. water. 

20 -r 28.5 = .701 per cent. Answer. 

The theoretical power of a pump required to raise a 
certain quantity of water to a certain height without re- 
gard to friction of the parts of the pump and of water in 
the pipes will be the same as the power of that quantity 
of water falling from that height But the friction in 
the parts of the pump and of the water in the pipes will 
average about \ of the total power of the water, or the 
power of the pump will be 1| times the total power of 
the water raised ; and, as the effective power of the water 
is but about 80 per cent, of the total power, the effective 
power of the water will average but about f the power 
required to raise it. In two tests of large pumping en- 
gines for citieSj the total power raised by one engine was 
78 per cent, of the indicated power required to raise it, 
while the other showed an exceptionally high rate, — 
namely, 96 per cent. 

To FIND H. P. FOR Pumps, Friction \ of Total 
H. P. REQUIRED. — Rules : Multiply the pounds of water 



THE engineer's EPITOME. 113 

pumped per minute by the pressure pumped against, in- 
eluding vacuum in suction pipe, if any, and by .00072; 
or multiply the cubic feet per minute by the pressure 
and by .0033 ; or, for head, multiply the pounds per 
minute in feet by head pumped against in feet, including 
height of suction pipe between the surface of water and 
pump, and by .000037 ; or multiply the gallons per min- 
ute by head and by .00033 ; or multiply the cubic feet 
per minute by head and by .0024. 

Examples. — What power of engine is required to raise 
1000 gallons of water 20 feet to pump, and force it against 
60 pounds' pressure ? 

20 X .432 = 8.64 

8.64 + 60 = 68.64 = total pressure. 

1000X68.64 = 68640 

.00072 X 88640 = 49.92 H. P. Answer. 

What engine power is required to raise 600 cubic feet 
of water 140 feet high above pump, there being 20 feet 
from surface of water to pump? 

140 + 20 = 160 = total head. 

600X160 = 96000 

.0024 X 9600j0 = 230.4 H. P. 

To FIND THE Diameter of Cylinder for a Pump. — 
Rule; Multiply the number of pounds of water dis- 
charged per minute by 3.5, or the gallons by 30, or the 
cubic feet by 222. Divide the product by the distance 
the pump piston moves the volume of water in the pum.p 
cylinder per minute, and extract the square root of the 
quotient. 

Formula (for gallons) : — 



v/: 



30 g 



in which L indicates feet per minute of pump piston for 



114 THE engineer's EPITOME. 

double acting pump, or 2 feet per minute for single act- 
ing pump. 

Example. — The stroke of a pump is 15 inches, double 
acting. What is the diameter of the cylinder to deliver 
500 gallons per minute? 120 strokes per minute? 
Piston motion equals 1| x 60 x 2 = 150 feet per minute. 
Then 

30 X 500= 15000 

15000-1- 150=100 

-i/ioo = 10 — diameter of cylinder. Answer. 

This rule gives the contents of pump cylinder about \ 
greater than the quantity actually passing through the 
pump. 

The number of gallons delivered per minute equals 
the feet per minute of piston multiplied by the square of 
diameter of pump cylinder divided by 30. The diameter 
of suction pipe should be such that the water will flow 
to and follow the pump piston, and fill the cylinder at 
its maximum velocity of stroke. The velocity of water 
in the suction pipe will vary with the amount of vacuum 
produced in the pump and with the height of pump 
above the surface of the water. The amount of vacuum 
in inches, shown by the gauge, multiplied by 1^ will 
equal the height the water will rise in the suction pipe ; 
and the velocity of flow in the pipe will be that due to 
difference of height of pump above the surface of the 
water and that due to vacuum in the pipe. 

Rule : For theoretical velocity of water in suction pipe 
multiply the square root of 1^ times the vacuum, in 
inches, by gauge in the pump, minus the height of pump 
above the surface of the water, by 8. 

Example. — The vacuum at the pump is 25 inches. 
The pump raises the water 18 feet. What is the velocity 
of water to the pump ? 



THE engineer's EPITOME. 115 



li X 25 = 


= 28.125 




28.125 — 


18 = 10.125 




-1,10.125 


==3.18 




8 X 3.18 


= 25.44 feet. 


Ansiver. 



To FIND THE Proper Diameter of Suction Pipe. 
Rule : Multiply the square of the diameter of pump 
cylinder, in inches, by the stroke, in feet, per minute. 
Divide the product by 400 times the velocity of the 
water in the pipe, and extract the square root of the 
quotient. 

^ -, / d- s 

Formula : V / 

V 400 V 

Example. — The diameter of the pump cylinder is 

6 inches, double acting, 80 strokes per minute ; stroke, 

16 inches; vacuum in pump, 20 inches; pump, 15 feet 

below surface of water. What should be the minimum 

diameter of suction pipe ? 

1^X20 = 22.5 
22.5 — 15 = 7.5 
-^'7^5"= 2.73 

400 X 2.73 = 1092 

SOX if =106.6 
36 X 106.6 = 3837.6 
3837 6 -f- 1096 = 3.5 

-yj3.5 = 1.87 = diameter. Aiiswer. 

This rule gives about one-fifth more water through the 
pipe than the capacity of the pump. 

In centrifugal pumps the quantity of water discharged, 
in cubic feet, equals the square of the diameter of the 
pipe, in feet, multiplied by the square root of the total 
height through which the water is raised and by 6.2. 
The height to which the water is raised depends on the 
velocity of the circumference of the pump, every certain 
velocity corresponding to a certain height. 



116 THE engineer's EPITOME. 

To FIND THE Height, divide the velocity of the cir- 
cumference by 500, subtract 1.1, and square the re- 
mainder. 

To FIND THE Velocity of Circumference, multi- 
ply the square root of the height by 500. To the product 
add 550. 

For raising large quantities of water to a small height 
centrifugal pumps are very convenient; but their aver- 
age efficiency falls much below the reciprocating pump, 
being, on a test for one maker, .523 of the power, and 
for another maker .603 of the power. 



THE ENGIXEER'S EPITOME. 117 



BELTS. 



The friction of a belt on a pulley varies as the tension 
and area of contact, independent of surface area ; i.e., if 
one pulley has twice the circumference of the other, the 
friction is the same for the same area of contact, al- 
though the surface in the last case is twice that of the 
other. The tension on a belt, when still, depends upon 
the amount of friction between the belt and pulley. If 
the coefficient of friction is 40 per cent., or the difference 
in tension between the working and slack sides of the 
belt is 40 per cent, of the working side, the tension on 
the sides of the belt, when still, is four-fifths of the 
tension on the working side. This working tension for 
a single leather belt of good quality is given as 67 
pounds per inch of width ; double thickness belts, as 108 
pounds. Belts are sometimes made double at the edges, 
when the tension can be anywhere from 67 to 108 
pounds per inch in width, depending upon the extent to 
which they are doubled. Rubber belts will convey about 
one-fifth more power than leather, but are not so dura- 
ble. The difference between tight and slack sides, 
divided by the tension on the tight, or working, side, will 
give the coefficient of friction. This is generally consid- 
ered as .42. One-half the difference in tension of the 
sides subtracted from tight side will give the tension on 
sides when the belt is still. 

Example. — The working strain per inch in width of 
belt is 70 pounds ; slack side, 30 pounds. What is the 



118 THE engineer's EPITOME. 

coefficient of friction, and what is the tension on the sides 
of belt when still ? 

70 — 30 = 40 

40 -r 70 = .57 = coefficient of friction. 

70 — \o = 55 pounds. 

The Working Strain on Tight Side of Belt 
AND Coefficient of Friction given, to find the 
Tension at which the Belt is to be laced. — 
Bule : Multiply the working strain by coefficient, and 
subtract one-half the product from the working strain. 

The Tension at which the Belt is laced and 
Coefficient of Friction given, to find the Work- 
ing Strain. — Rule: Divide twice the tension at which 
the belt is laced by 2 minus the coefficient of friction. 

To find the Speed of Belt per Minute in Feet. — 
Eule : Multiply the diameter of either pulley, in inches 
by revolutions per minute and by .26. 

Example. — The diameter of pulley is 20 inches ; revo- 
lution, 300 per minute. What is the speed of belt per 
minute in feet? 

300 X 20 = 6000 

.26 X 6000 = 1560 feet. Answer. 

To FIND THE Arc of Contact of Belt with 
Pulleys. — In open belts, the arc of contact of large 
pulley equals 360° minus the arc of contact of small 
pulley. 

To FIND the Arc of Contact of Belt (Open 
Belt and Small Pulley). — Rule : Subtract the differ- 
ence of radii from distance ]5fetween centres. Divide the 
remainder by distance between centres. Add together 
162 times the square \ of the quotient, 3 times the quo- 



^O^i-oi-«taje. K.OBT- 



THE engineer's EPITOME. 119 

tient, and 15 times the square of the quotient. The sum 
equals the arc of contact of belt with pulley in degrees. 
Formula : 

= m (162 yJm -|- 3 m + 15 m^^ = arc. 

The arc of contact in crossed belts is the same for 
both pulleys, and is always greater than 180°. 

To FIND THE Arc of Contact of Crossed Belts 
— Rule : Subtract the sum of radii from distance between 
centres. Divide the remainder by distance between cen- 
tres. Add together 162 times the square root of the quo- 
tient, 3 times the quotient, and 15 times the square of 
the quotient. Subtract the sum from 360°. The re- 
mainder equals the arc of contact of belt with both 
pulleys. 

Formula: 

~ ^ "^ ^^^ = m 360 — (162 sjm-^ 3 m + 15 m^) = arc. 

Examples. — The diameters of the pulleys are 40 and 
10 inches ; distance between centres, 12 feet, open belt. 
What is the arc of contact for both pulleys ? 

12 feet equals 144 inches. Then 

40 ^ 2 = 20 = radius. 

10-7-2 = 5 = radius. 

20 — 5 = 15 

144—15 = 129 

129J- 144 = .895 

-J^SgS = .946 

162 X 946 = 153.25 

3 X. 895 = 2.685 

.89.52 ^ 801 

15 X -801 = 12.015 

153.25 + 2.685 + 12.015 = 167.95° = contact of small pulley. 

360 — 167.95 = 192.05° = contact large pulley. 

The diameters of the pulleys are 16 and 4 inches; 



120 THE engineer's EPITOME. 

distance between centres, 4 feet, crossed belt. What is 
the arc of contact of belt with the pulleys ? 

4 feet = 48 inches. 

8 + 2 = 10 

48 — 10 = 38 

38-^48 = .791 

^."791=. 889 

162 X .889 = 144.018 

3 X .791 = 2.373 

2.373 + 144.018 = 146.391 

.7922= 6256 

15 X. 6256 =9.384 

146.391 +9.384 = 155.77 

360 — 155. 77 = 204.23 = arc of contact. 

While the H. P. of a belt varies as the arc of contact, 
it does not vary as directly as that arc ; i.e., when the 
arc of contact of belt is 60°, the power of the belt is 
about 2^ of the total power of whole circumference ; at 
120°, about ^f ; at 180°, about if; while the power for 
the whole circumference is less than 2^ times the power 
at 60°. 

To FIND THE H. P. OF A Belt. — Rule: Multiply the 
working strain per inch, by width in inches, by velocity 
in feet per minute, and divide the product by 448S„ 
divided by arc of contact (small pulley, open belt) plus 
17, plus gijj of arc of contact. Point off three figure 
from the quotient. 

T(wf) 



Formula: .001- 



4488 



a + lT + eVa 
in which T indicates working strain on belt; w, width in 
inches ; f, feet per minute ; a, arc of contact in degrees- 
Example. — The tension on working side of belt is 75 
pounds ; width, 10 inches ; speed, 2500 feet per minute; 
arc of contact, 150°. What is the H. P. transmitted? 



THE engineer's EPITOME. 121 

10X75 = 750 

2500X750= 1875000 

4488 -f- 150 = 29.92 

29.92 + 17 = 46.92 

150 -^ 60 = 2.5 

46.92 + 2.5 = 49.42 

1875 -f 49.42 = 37.94 H. P. Ansiver. 

As the position in which principal belts are generally- 
placed is such that the arc of contact will not vary to 
any considerable extent from 180°, and as the H. P. of 
belts varies but slightly for a considerable variation of 
arc of contact from 180°, simple approximate rules can 
be given for H. P. of belts. 

For finding the H. P. of Single Belts. — Rule: 
Multiply the width in inches by feet per minute. Point 
off two figures from the right hand, and divide by 7. Or, 
if the tension on tight side of belt is given, multiply the 
tension by the width, by the feet per minute. Point off 
three figures, and divide by 45. 

For double belts proceed as above, except in the first 
division divide by 4, and in the second division by 28. 

Examples. — Width of single belt, 15 inches; velocity, 
1800 feet per minute. What is the H. P. ? 

15 X 1800 = 27000 

27000 -i- 700 = 38.57 H. P. Answer. 

The width of a double belt is 36 inches ; velocity, 4000 
feet per minute ; tension on working side, 110 pounds per 
inch in width. What is the H. P. transmitted? 

36X110 = 3960 

4X3960= 15840 

15840 -f 28 = 565.71 H. P. Answer. 

To find Length of Belts (for Open Belt). — Rule : 
Multiply the sum of the radii by 3.14. To product add 



122 THE engineer's epitome. 

twice the distance between the centres, and the square 
of the difference between the radii divided by the dis- 
tance between the centres. 

Formula : 3.14 (R + r) + 2 D + ^^~^^' 

Rule for crossed belts : Multiply the sum of the radii 
by 3.14. To the product add twice the distance between 
the centres and the square of the sum of the radii divided 
by distance between the centres. 

Formula : 3.14 (R -|- r)-|- 2 D + (?dlll' 

These rules give very correct results, except in very ex- 
treme cases where the difference in radii is very great 
and the distance between the centres very small. 

Example. — (Illustrating rule for open belts.) The di- 
ameters are 30 and 24 inches ; distance between centres, 
10 feet. What is the length of belt required V 

30-^-2 = 15 = radius. 

24 -f 2 = 12 = radius. 

15 + 12 = 27 

3.14X27 = 84.78 

1-0 feet = 120 inches. 

2X120 = 240 

84.78 + 240 = 324.78 

15 — 12 = 3 

3^ = 9 

9-M20 = .07 

324.78 + .07 = 324.85 inches. 

324.85 -M2 = 27.07 feet. Answer. 

Example. — (Illustrating rule for crossed belts.) The 
diameters are 18 and 6 inches; distance between centres, 
40 inches. What is the length of belt ? 

18 -1-2 = 9 = radius. 
6 -f 2 = 3 = radius. 
3.14 X 12 = 37.68 
2 X 40 = 80 



THE engineer's EPITOME. 123 

37.68 + 80=117.68 
9 + 3 = 12 
122=144 
144-^-40 = 3.6 

117.68 + 3.6=- 121.28 inches. Answer. 

The following formulas will give the length of belt in 
any case, but the processes are much more difficult : — 

Formula for open belt : — 

E _ r 

. = m ( (2 — .13 m) ^' 1 — m (1 — m) ) D -f 6.28 R 

Formula for crossed belt : — 

?-t^ = m ( (2 — .13 m) ^ 1 — m (1 — m) + 6.28 m) D 

For cone pulleys, crossed belts, make the sum of the 
diameters of opposite pulleys all equal. For cone 
pulleys, open belt, fix on the proper size of pulleys on 
one cone, and the smallest pulleys on the opposite or 
required cone. Assume other pulleys for this required 
cone as for a cross belt, so that the sum of the diam- 
eters of opposite pulleys are equal to the sum of the 
diameters of the largest and smallest opposite pulleys. 
From the square of the difference of radii of the largest 
and smallest pulleys on opposite cones subtract the 
square of the difference of radii of any other set of 
pulleys given and assumed, and divide the difference 
by 3.11 times the distance between centres plus or 
minus twice the difference of radii between the given 
and assumed pulleys, — plus, if the assumed pulley is 
greater than the given, minus if less. Add the quo- 
tient to the assumed pulley. 
Formula : 

/R y.y2 -£2 

—, = correction to be added to assumed 

3.14 D± 2 E 

pulley, in which R and r indicate radii of largest and 



124 THE en'gineer's epitome. 

smallest pulleys on opposite cones, and E the difference 
of radii between any other opposite pulleys. 

While the friction on two pulleys remains the same 
for the same arc of contact, the power transmitted will 
increase as the diameters of the pulleys increase, owing 
to the increased velocity of belt. The revolutions, how- 
ever, remain the same ; that is, if two pulleys are so 
placed as to make an arc of contact of 150 with the 
smallest pulley. The diameters of the pulleys should 
be doubled, and placed so as to maintain the same arc 
of contact, the revolutions and width of belt remaining 
the same. The H. P. transmitted would be doubled, be- 
cause of the double velocity of the belt, and not because 
of the greater area of contact. 

A leather belt should not be allowed to become dry 
and stiff, but should be kept pliant by the application 
of some softening material. 

Rule for H. P. of rope belts : Multiply the square of 
the diameter of the rope in inches by velocity in feet 
per minute, and by .003: the product equals the H. P. 
The diameter of the smallest pulley should be 30 times 
the diameter of the rope. Average breaking strength of 
the rope should be 8000 times the square of the diame- 
ter ; average weight per foot, in length, ^ the square of 
the diameter. 



THE engineer's EPITOME. 125 



GOVERNORS. 



In pendulum governors the height formed from a 
line passing through the centre of gravity of balls to 
the intersection of the lines passing through the points 
of suspension, depends upon the number of revolutions. 
The same number of revolutions produce the same 
height without regard to the weight of balls or length 
of arms. If the point of suspension is in the axis of 
revolution, the balls will be in the same horizontal plane, 
whatever the length of arms or weight of balls ; but the 
point of suspension is very often outside the axis of 
revolution, making the computation of position of balls 
very difficult. To make the governor more sensitive, 
a centre weight is frequently connected by links to the 
balls, the effect of which is to increase the height in 
proportion, as 1 is to twice the central weight plus the 
weight of the balls, divided by the weight of the balls, 
thereby producing a much quicker and greater opening 
and closing of the governor valve with the same regula- 
tion of revolution. Governors are made to run at a 
certain number of revolutions, which is generally found 
stamped upon them by the maker. The speed of the 
engine is changed by giving different diameters to the 
governor pulleys, the revolutions of the governor being 
unaltered. 

To CHANGE THE SpEED OF THE EnGINE WHEN THE 

Pulley or Gear for driving the Governor is 



126 THE engineer's epitome. 

FIXED ON THE Shaft. — Rule : Multiply the required 
number of revolutions of engine by the diameter of 
pulley on engine shaft, and divide the product by the 
number of revolutions of governor. The quotient is 
the diameter of pulley for governor shaft. 

Example. — The engine is running at 80 revolutions; 
pulley on shaft, 10 inches ; governor, 100 revolutions. 
It is desired to raise the speed of the engine to 120 
revolutions. What diameter of pulley is required on 
the governor? 

10 X 120= 1200 

1200 -f- 100 = 12 inches. Answer. 

To FIND Height of Balls. — Rule : Divide 187.5 
by the number of revolutions of governor, and square 
the quotient. 

Example. — The number of revolutions of governor 
are 60. What is the height of apex of arms above the 
plane of centre of balls ? 

187.5-1-60 = 3.12 

3.122= 9.73 ^ height. Answer. 

To FIND THE Centrifugal Force. — Rule: Multiply 
the weight by the square of the number of revolutions 
per minute, by the diameter of centre of gravity of 
weight, in feet, and by .00017. 

Formula : .00017 (R^ D) w 

The diameter of the centre of gravity or gyration in 
a fly wheel will nearly correspond to the diameter of 
the wheel, unless the wheel is small and has a very thick 
rim. 

Example. — The weight of the fly wheel is 1200 pounds, 
revolutions 120 per minute, diameter 8 feet. What is 
the centrifugal force? 



THE p:ngineer's epitome. 127 

120- =14400 

14400 X 1 200 = 1 7280000 

8 X 17280000 = 138240000 

.00017 X 138240000 = 23500.8 pounds. Ansicer. 

Calling the tensile strength of cast iron 20000 per 
square inch, the centrifugal force will be about y^y of the 
bursting strain on the wheel, there being about 12 
square inches area in the rim. 

To FIND Weight of Fly Wheel. — Rule : Multiply 
240 times the effective pressure in the cylinder by the 
square of its diameter in inches, and by the stroke in 
feet. Divide the product by diameter of wheel mul- 
tiplied by revolutions per minute. 

Following is a general formula : — 
s (12000 d^) 
^R 
This formula gives a very uniform speed, if the speed is 
not required to be very regular. A wheel of less weight 
can be used, thereby decreasing the weight on the 
bearings. 



128 THE engi^-eer's epitome. 



PIPE SURFACE FOR HEATING. 



In heating buildings by steam or water, from .7 to 
1.2 units of heat will be required to be transmitted for 
every square foot of surface of outside wall, depending 
upon the thickness of the wall and the material of which 
it is composed. From a 12-inch brick to a thin wooden 
wall, the transmission of heat from pipes is as the 
square of the difference of heat in the pipes and the 
surrounding air ; and the transmission for one square 
foot of pipe is one one-hundredth of this square. 

To FIND the Square Feet of Pipe Surface to 
heat a Building. — Rule : Multiply the outside surface 
of the walls by from 70 to J.2.5, depending upon the 
thickness and material of the wall, by the difference 
between the temperature required inside and the outside 
temperature. (In this latitude, the outside temperature 
is taken at zero.) Divide the product by the square of 
the difference in temperature between the pipes and that 
required in the building. 

The diameter of a main steam pipe should be -^-^ of the 
square root of the pipe surface. 

To WEIGH a Shaft, Long Irregular Casting, or 
Pillar, or Long Stick of Timber, with a Small 
Steel Rod. — Rule: Balance the article to be weighed 
on a roller, and mark the centre of gravity. Move it 
along the roller until the heavy end is about equal to 
the capacity of the steelyard, mark the position of roller, 



THE EXGIXEER S EPITOME. 



129 




and find the weight of the heavy end. Divide the dis- 
tance from roller or support to heavy end, by the 
distance moved from centre of 
gravity. Multiply the quotient 
by weight of heavy end. 

Example. — An unequal stick 
of timber is moved 2 feet from 
its centre of gravity, and is sup- 
ported 15 feet from heavy end. 
The weight of heavy end is 75 
pounds. What is the whole 
weight of the timber ? 

15-^-2= 7.5 

7.5 X 75 = 562.5 pounds. Ansicer. 

To FIND THE Cubical Cox- 
tents OF A Wedge cut from a Cylinder. — Rule : 
Multiply the square of the diameter of the cylinder by 
length of wedge and by .452. 

Formula : .452 d-^L 



To find the Cubic Feet in a Round Shaft or 
Stick of Timber. — Rule : Multiply the square of the 
girth in inches by the length in feet. Divide the prod- 
uct by 18, and cut off two figures. 

If a log, and the board measure is required, divide the 
product by 15, and cut off three figures. If the shaft 
or log is tapering, take the girth from the large end. 
Where the diameter of shaft is given, multiply the square 
of the diameter in inches by length in feet. Divide by 
18.3, and cut off one figure. 

Example. — The length of a propeller shaft is 25 feet ; 
girth, 30 inches. How many cubic feet of iron does it 
contain, and what is the weight ? 

A cubic foot of wrought iron weighs 480 x^ounds. 



130 THE engineer's EPITOME. 

302 ^ 900 

900 X 25 = 22500 

22500^ 18 = 1250 

Cutting off two figures gives 12.5 cubic feet. 

12.5 X 480 = 6000 pounds. 

To FIND THE Cubical Contents of Segment of 
Sphere. — Rule: From three times the radius of the 
sphere multiplied by the height of the segment subtract 
the square of the height of segment. Multiply the re- 
mainder by 1.0572 times the height of segment. 

Formula : 1.0572 h (3 rh) — h2 

To FIND THE Cubical Contents of Zone between 
the Segment and Diameter of the Sphere: — 

Formula : 1.0572 (2 r^ _ h^) 3 r — h 
in which r indicates radius of sphere, and h height of 
segment, as before. 

Example. — The diameter of sphere is 40 inches ; 
height of segment, 6 inches. Find the cubical contents 
of the segment. 

40 -f 2 = 20 = radius. 

20 X 6 = 120 

3X 120 =-360 

62 = 36 

360 — 36 = 324 

1.0572 X 6 = 6.34 

6.34 X 324 = 2054.16 cubic inches. Answer. 

To FIND THE Area of a Segment of a Circle. — 
Rule : Divide the height of the segment by the diameter 
of the circle. Multiply 1.3312 minus .381 times the quo- 
tient, plus .12 times the square of the quotient, by the 
product of the quotient multiplied by the square root of 
the quotient. Multiply the last product by the square of 
the diameter of the circle. 

Formula: — = m D^ (1.3312 — (.381 + .12 m2) m^ m) 



THE engineer's EPITOME. 131 

Example. — The diameter of a boiler is 8 feet. The 
steam space above the water line is 2 feet. What is the 

area ? 

2-i-8 = .25 

.381 X .25 = .0952 

.25- = .0625 

.12 X .0625 = .0075 

.09.52 4- .0075 = .1027 

1.3312 — .1027 = 1.2285 

^."25"= .5 

.5X .25 = .125 

.125 X 1.2285 = .1535 

8-- = 64 

64 X .1535 = 9.82 square feet. Ayiswer. 

To FIND THE Sine of an Angle Less than 50°. — 
Formula : 

.0000000808 a (.3600 ^ 

^ (19786-f .000099 a^) — . 07 a^' 

a indicates angle in minutes. 

For Sine of Angle between 50° and 90°. — 
Formula : 

.0000000808 a (3600 ) 

"^ (19902 + .000106 a2) — .13 a^ 

Example. — Find sine of angle 25°. 

25° equal 1500 minutes. 

1500- = 2250000 

.07 X 1500 = 105 

.000099 X 2250000 = 222.75' 

19786 + 222.75 = 20008.75 

20008.75 — 105 = 19903.75 

2250000^ 19903.75 = 113.04 

3600 — 113.04 = .3486.96 

1500 X 3486.96 = 5230440 

.0000000808 X 5230440 = .4226 = Avf^wer. 

To find Closely Approximate Length of Cycloi- 
DAL Circle, ^ the Length. — Formula: — 

2r(l.57 + .43(-^>) 



132 THE engineer's epitome. 

when the generating point is within the radius r of the 
rolling circle, and as follows when the generating point 
is outside the rolling circle : — 

2g(l.57+.43(-I-f) 

When the points g and r coincide, the true cycloid 
is formed. 

To FIND THE Length of Any Part of the True 
Cycloid Curve. — Formula : — 
1-^/87^ 
in which r indicates radius of rolliog circle ; x, height 
from apex or top of curve ; 1, length of curve from the 
apex. 

A wheel 48 inches in diameter, with the generating 
point in its circumference on the base, is rolled until the 
generating point is one foot above the base or surface. 
What is the length of the curve described by the point ? 

2X2 = 4 

4 X 24 = 96 = length of curve to apex. 

48 — 12 = 36 = height from point to apex. 

8 X 24 = 192 

36 X 192 = 6912 

-^6912 = 83.13 

96 — 83.13 = 12.87 inches. Answer. 



To FIND Elevation of Outer Rail in Railroad 

Curves. — Formula : — 

.8 X gauge x (miles per hour)^ . . . , 

— ,.^ 1- ^^ . : — ; — r = elevation m inches. 

radius of curve in inches 



THE engineer's EPITOME. 133 

To FIND Minimum Tractive Force of Locomo- 
tives. — Formula : — 

= power for 1 pound steam. 

in which D indicates diameter of cylinder; s, stroke; 
w, diameter of driving wheel. 

The average tractive force is found by the following 
formula : — 

wl 
in which 1 indicates length of connecting rod. 

Adhesion to the rails is estimated in various ways. 
American practice is i total weight of locomotive, which 
equals the adhesive friction between wheel and rail. 
This depends much upon the condition of the rail. If 
the traction force is greater than the adhesive, the wheels 
will slip. 



134 THE engineer's epitome. 



SALT IN ^A^ATER. 



Multiply the number of ounces of salt in one gallon 
of water by .235, and add the product to 212. The sum 
equals the boiling point due to saltness. 

Multiply the difference in height of barometer and 30 
inches, in hundredths of an inch, by .016. Add or sub- 
tract the product to or from 212, which gives, approxi- 
mately, the temperature of boiling point due to pressure. 

Five ounces of salt to the gallon is common sea- 
water, = Jg by weight of salt. 60 ounces of salt to the 
gallon equals |^| weight of salt, and the water is fully 
saturated. 

To FIND Boiling Point due to Saltness and Dif- 
ference IN Barometer Pressure. — Rule: Add to or 
subtract from the temperature due to saltness .016 times 
the difference, in hundredths of an inch, between barome- 
ter and 30, as the barometer is greater or less than 30 
inches. 

Nine times the difference in barometer readings in 
hundredths of an inch equals approximate height in feet, 
and 550 times the difference in temperature of boiling 
water equals approximate height in feet. 

To FIND Approximate Height by a Barometer. — 
Rule : Divide reading of barometer at lowest station by 
reading at highest station. Multiply the quotient by 
2400 and the square of the quotient by 180. Add to the 
sum of the products 51000. Multiply this sum by the 



THE engineer's EPITOME. 135 

quotient of the difference of the readings divided by the 
sum of the readings. 

To correct for temperature, add or subtract from the 
height found above, the part of the height multiplied by 
the difference between the average temperature at the 
stations and 70°, as the temperature is greater or less 
than 70°. 

The weight of a cubic foot of dry air equals 2.709 
times the absolute pressure divided by 461.2 plus the 
temperature. If the air is changed from one tempera- 
ture to another, the volume remaining the same, the 
pressure is changed too : — 

F 46P + T 
4612-ft 
in which P indicates the absolute pressures, and T and t 
the temperatures. 



THE ENGINEER S EPITOME. 



THE MASON REDUCING VALVE. 

This valve is designed to reduce and maintain an 
even steam or air pressure regardless of the initial press- 
ure. It will automatically reduce boiler pressure for 
steam-heating coils, dry rooms, paper-making machin- 
ery, slashers, dye kettles, and all places where it is 
desirable to use lower pressure than that of the boiler. 
The dash-pot, which immediately fills with condensa- 
tion, prevents all chattering or pounding, and requires 
ro attention. No extra lock-up attachment is needed, 
as the pressure is regulated by a key, which the en- 
gineer retains. The sizes, up to and including 2-inch, 
are made of the best composition, and above that, of 
cast iron, with composition linings. In the larger sizes, 
the composition lining is hung up in the valve, leaving 
a space between the iron and composition for the un- 
equal expansion of the metals. Thus there is no possibility of the piston 
sticking when the valve is heated. The area of the passage from the high 
to the low pressure side of the valve is equal, when open, to thf full area of 
the pipe, so that a low pressure of the system, almost equal to the initial 
high pressure, may be carried. This valve will maintain an even steam or 
air pressure as low as one pound if necessary. This valve is warranted to do 
all that is claimed for it, and will be sent to responsible parties on 30 days' 
trial. We make an iron valve for use with ammonia in ice-making machin- 
ery. 




SIZE PIPE, 

1/2 inch 



PRICE LIST. PRICE. 

|i5-oo 

18.00 

22.00 
28.00 

3500 

4400 

S7-00 

72.00 
100.00 

135-00 

180.00 
250.00 



Manufactured by the MASON REGULATOR CO., Boston, Mass. 



THE ENGINEER S EPITOME. 



THE MASON PUMP PRESSURE REGULATOR. 

For Fire, Tank, Elevator, Air and Water Works 
Pumps, or any class of Pumping Machinery where 
it is necessary to maintain a constant pressure. The 
Mason Pressure Regulator is made upon an entirely new 
principle, the advantage of which is that the steam itself 
is made both to open and shut the valve. The Regulator 
may be instantly adjusted to any pressure desired by 
simply turning the key, as shown in the cut. 

The especial feature of this regulator is that the press- 
ure chamber into which the water enters is entirely 
removed and separate from the steam and all working 
parts. 

The long cylinder at the bottom of the regulator is a 
dash-pot, the piston of which is connected with the main 
valve of regulator, thereby preventing sudden and violent 
"jumping" of the pump when the pressure suddenly changes. 

For automatic fire sprinkler service they have been found especially valua- 
ble, as the valve is thrown wide open, immediately the slightest drop in pres- 
sure occurs. On application we can refer parties where they are in use. 

These regulators are made in all the pipe sizes ; those up to and including 
2-inch, of the best steam metal; the larger sizes, of cast iron, lined with 
steam metal. The springs are made of the finest tool steel, tempered. 




; PIPE. 
, inch 



PRICE LIST. 



PRICE. 

I17.OO 
20.00 
25.00 
30.00 
42.00 

SS-oo 
68.00 



5.00 



Manufactured by the MASON REGULATOR CO., Boston, Mass. 



THE ENGINEER'S EPITOME. 



THE MASON LOCOMOTIVE REDUCING 
VALVE. 

This reducing valve has been in use by the leading 
steam car-heating companies for the last three years. 
During that time over 3,000 have been placed on 
nearly every railroad which has adopted steam heat. 
One feature which will commend itself to every engineer 
is the self-locking device, which enables the valve to be 
set for any pressure, and automatically locked. In the 
manufacture of this valve there is no attempt made to 
save on stock in order to cheapen the price. Every 
valve is made of the best steam metal, and will not 
steam-cut at a pressure of 180 pounds. As we supply a 
e number of car-heating and railroad companies, 
each of whom use different connections, we are pre- 
pared to furnish the sizes as follows : — 




The 



inch size. 



iK 



PRICE. 

$18.00 
22.00 
28. GO 



THE MASON AIR-BRAKE PUMP REGULATOR. 

This is constructed on the same principle as 
our pump pressure regulator, but with a few 
small changes, so as to allow of its application 
to air-brake pumps without changing the fit- 
tings on the locomotive. It is only necessary 
to place these in the steam pipe leading to the 
pump, and connect the air pressure pipe with 
the train service pipe. As the air chamber is 
entirely separate from the steam and working 
parts, any particles of dirt which may get into 
the main service pipe cannot get into the work- 
ing parts of the regulator and thence into the 
pump, neither can the steam and air pressure 
have access to each other. 

We will send one on trial to any railroad. 
Weight, with couplings, 8J^ pounds. 

Price, 515 net. 



Manufactured by the MASON REGULATOR CO., Boston, Mass. 



THE ENGINEER S EPITOME. 




THE MASON PUMP GOVERNOR. 

The Mason Pump Governor 
is to the direct acting steam 
pump what the ordinary ball 
governor is to the steam engine. 
It attaches directly to the pis- 
ton rod of the pump, and oper- 
ates a balanced valve placed in 
the steam pipe, thereby exactly 
weighing the amount of steam 
to the needs of the pump, and 
economizing the same. By 
using the Mason Governor, you 
can instantly set or change your 
pump to any required speed, 
which will be maintained in spite 
of variation in load or steam 
FIG. I. pressure. As all the working 

parts of the Governor are immersed in oil, the wear is reduced to a minimum. 
It is largely used on vacuum pumps, deep-well pumps, water-works pumps, 
ice machines, and all classes of pumps requiring a uniform stroke. 

For duplex pumps, the Gover- 
I nor is fitted with a special valve, 
I which holds the main piston sta- 
tionary during the momentary 
pause of the pump piston, from 
which the motion is taken. 

For large duplex pumping en- 
gines a special Governor is made, 
fitted with double parts, and con- 
FiG. 2. nected with both piston rods, thus 

insuring perfect regulation for every portion of each revolution of the engine. 
Figure i shows the Governor unattached. 
Figure 2 shows the Governor connected with balanced valve. 







PRICE LIST OF GOVERNOR, WITHOUT VALVE. 

We make three sizes of these Governors : — 



2 No. I goes with Y^ in. steam valve to 2 in. inc., 

<'! ^ a <'<' ''/2 ;; ;; ;; " 4 " " 

3 5 and upward, 



PRICE. 

^45.00 
60.00 



Manufactured by the MASON REGULATOR CO., Boston, Mass. 



THE ENGINEER S EPITOME. 

THE MASON STEAM DAMPER REGULATOR 

Is made in accordance with an entirely novel princi- ' 
pie in the construction of damper regulators : the steam 
is admitted from the boiler, through a small strainer, as shown 
in the cut, under a diaphragm which is attached to an auxiliary 
valve, which either lets in the steam or excludes it from the 
piston which works in the cylinder. This piston, being forced 
up or down by the steam pressure, turns a wheel, from which 
runs the chain which is connected to the damper. 
By an ingenious arrangement, the damper is not -__ 
thrown entirely open by the slightest change of 
pressure in the boiler, but is gradually opened, as i® 
the pressure changes, thereby 
not allowing sudden and great 
variations of draft, whereby the 
coal is wasted. 

We manufacture only one size. 
The regulators are made of the 
best steam metal throughout. 
Each one is nickel-plated. Weight, thirty 
pounds. Price, $80, including bracket, 
weight, and twenty-five feet of chain. 

We manufacture our Lever and Balanced 

Valves, as shown on following page, in%'^^ 

two different styles. The piston valves 

are not perfectly steam tight. The bevel-seated valves are intended for use 

where it is required for the valve to shut absolutely steam tight. 

Style A, Balanced Valve Piston, with straight stem guide. 
" B, " '■ " " lever attachment. 

" C, " " Bevel-seated, with straight stem guide. 

" T>, " " " " " lever attachment. 




Manufactured by the MASON REGULATOR CO., Boston, Mass. 



THE ENGINEER S EPITOME. 



THE MASON BALANCED VALVE 




Is a double piston balanced valve, for use in connection with our speed gov- 
ernor, and for which we have found a ready sale in the many cases where 
such valves are used. The principal advantage of our valve is in the guide 
for the valve stem, which is cast on the bonnet, thereby keeping the valve 
stem in a direct line with the stuffing box. For this reason it is impossible 
for the valve stem to bend, and the packing seldom has to be renewed. The 
sizes up to 2-inch are m-ide of the best steam metal; above that, of cast 
iron, lined with steam metal. A knuckle-joint tapped for three-eighths rod 
is on the same stem of each valve. This saves expense n making connec- 
tions. 

THE MASON LEVER VALVE 




Is made essentially the same as the " Mason 
Balanced Valve," with the substitution of yoke 
and lever with weight attached, for the bonne 
and knuckle-joint of the balanced valve. The 
lever valve will be found useful in controlling 
the supply of water in tanks by attaching to a 
ball float. It can also be used to control steam 
pumps in tank service, by placing the valve in 
the steam supply pipe to the pump, and con- 
necting the lever with the ball float placed in 
the tank. There is no lost motion in any of 
the joints, a fact which those desirous of close 
regulation will appreciate. These valves can 
be furnished either as piston or seated valves. 
The latter style is steam tight. For prices, see next page. 



Manufactured by the MASON REGULATOR CO., Boston, Mass. 



THE ENGINEEE'S EPITOME. 



PRICE LIST OF BALANCED AND LEVER VALVES. 

SIZE VALVE. 

V2 inch, 



PRICE. 

?S-5o 
6.50 
8.00 
9.00 
10.50 
13.00 
19.50 
25.00 
37.00 



CABLE ADDRESS, " MASONICA," BOSTON. 



TELEGRAPH CODE. 



For quantity of goods wanted use manerals. 



Locomotive Reducing Valves, 
Pump Pressure Regulators, 
Lever Valves, 
Pump Governors, . 
Balanced Valves, . 
Reducing Valves, . 
Damper Regulators, 
Air Brake Regulators, 



"%. inch, 



Locotnotive 

Pressure 

Lever 

Governor 

Balance 

Reducer 

Damper 

Brake 

Madder 

Maggot 

Magnate 

Major 

Malay 

Malice 

Mangle 

Manly 

Marten 

Master 

Mastic 

Mattock 

Ship via Express, Worship 

" " Freight, Welkin 

Example. — Ship by express five 2-inch reducing valves, " Worship five 

Malice Reducers." 



Manufactured by the MASON REGULATOR CO., Boston, Mass. 



M 



N 



J 



\1/e manufacture 

Reducing Valves. 

Pressure Regulators. 

Pump Governors. 

Ai r Brake Regulators. 
Locomotive Reducing Valves. 

Elevator Pump Regulators. 
Lever Valves. 



TUB mason Regulator Go. 

10 CENTRAL STREET, 

BOSTON, MASS. 



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